ow 0 be 
1 x 


.f, 
; en SS 
ee epee A E 
ae 
ep es : 
x = 2 ee J Z je . -! i 
: “ace hs re ee ; ‘ rs 

os i ‘ =. ’ 


ee 
rez 
ne 
eG 


4a 
es 
eet S 
Se 
ab ote . 
th sok eee a eee 
: : 1: Spe teh f : 


= peers = ee 


: a 

a 

ve Tae ae 3 

ee ee ay bt Mer" 
ed ee 


-® psa 
et 
pa ee 


he 


; pare OF - 
4,% 
i * 


Sate’ ar 
vane ~ 





LIBRARY OF THE 
UNIVERSITY OF ILLINOIS 
AT URBANA-CHAMPAIGN 


owe 
De8et 











ey a 7) iy A 
‘ 7 ‘ 
” 
vi j 
i 
a 


= he, 








a 
> 


“rr 








| 


| 


as 


——----- Oo 
Fe ee 








ELEMENTS 


OF 


ALGHBR A: 


Peay DING STURMS’ THEOREM. 


TRANSLATED FROM THE FRENCH OF 


M. BOURDON. 


ADAPTED TO THEE COURSE OF MATHEMATICAL INSTRUCTION IN THE 
UNITED STATES, 


BY CHARLES DAVIES, LL.D. 


AUTHOR OF ARITHMETIC, ELEMENTARY ALGEBRA, ELEMENTARY GEOMETRY, 
PRACTICAL GEOMETRY, ELEMENTS OF SURVEYING, ELEMENTS OF 
DESCRIPTIVE AND ANALYTICAL GEOMETRY, ELEMENTS 
OF DIFFERENTIAL AND INTEGRAL CALCULUS, 

AND A TREATISE ON SHADES, SHAD- 

OWS, AND PERSPECTIVE. 


NEW YORK: 
PUBLISHED BY A. S. BARNES & CO. 


No. 51 JOHN STREET. 
1848. 


DAVIES’ 
COURSE OF MATHEMATICS, 


DAVIES’ FIRST LESSONS IN ARITHMETIC—For Beginners. 
DAVIES’ ARITHMETIC—Designed for the use of Academies and Schools. 
KEY ‘FO DAVIES’ ARITHMETIC. 


DAVIES’ UNIVERSITY ARITHMETIC—Embracing the Science of Num- 


bers and their numerous Applications. 


KEY TO DAVIES’ UNIVERSITY ARITHMETIC, 


DAVIES’ ELEMENTARY ALGEBRA—Being an introduction to the Sci- 
euce, and forming a connecting link between ARITHMETIC and ALGEBRA. 


KEY TO DAVIES’ ELEMENTARY ALGEBRA. 


DAVIES’ ELEMENTARY GEOMETRY.—This work embraces the ele- 
mentary principles of Geometry. The reasoning is plain and concise, but at the 
same time strictly rigorous. 


DAVIES’ ELEMENTS OF DRAWING AND MENSURATION — Ap- 
plied to the Mechanic Arts. = 


DAVIES’? BOURDON’S ALGEBRA—Including Srurm’s Tororem—Being 
an abridgment of the Work of M. Bournon, with the addition of practical examples. 


DAVIES’ LEGENDRE’S GEOMETRY anp TRIGONOMETRY—Being 
an abridgment of the work of M. Legendre, with the addition of a Treatise on MEn- 
SURATION OF PLANES AND SOLIDS, and a Table of LOGARITHMS and LOGARITHMIC 
SiNEs. 


DAVIES’ SURVEYING—With a description and plates of the THropotire, 
Compass, PLANE-TABLE, and LEVEL; also, Maps of the TopoGRaPHicaL SiGNs 
adopted by the Engineer Department—an explanation of the method of surveying 
the Public Lands, and an Elementary Treatise on NAVIGATION. 


DAVIES’ ANALYTICAL GEOMETRY — Embracing the Equations or 

THE PoINT AND STRAIGHT L1inE—of the:Contc Sections—of the Link AND PLANK 

In Space; also, the discussion of the GENERAL EQUATION of the second degree, and 
of SURFACES of the second order. 


DAVIES’? DESCRIPTIVE GEOMETRY—With its application to SpHrr- 


ICAL PROJECTIONS. 
DAVIES’ SHADOWS anv LINEAR PERSPECTIVE, 
DAVIES’ DIFFERENTIAL ann INTEGRAL CALCULUS. 





Entered, according to Act of Congress, in the year 1844, by CuarLeEs DavigEs, in the Clerk’s 
Office of the District Court of the United States, in and for the Southern District of 
New York. 


\eu & | MA on 4X 


PREFACE 


i 

Tue Treatise on Algebra, by M. Bourdon, is a work of 
singular excellence and merit. In France, it is one of the 
leading text books. Shortly after its first publication, it passed 
through several editions, and has formed the basis of every 
subsequent work on the subject of Algebra. 

The original work is, however, a full and complete treatise 
on the subject of Algebra, the later editions containing about 
eight hundred pages octavo. The time which is given to 
the study of Algebra, in this country, even in those semin- 
aries where the course of mathematics is the fullest, is too 
short to accomplish so voluminous a work, and hence it has 
been found necessary either to modify it, or to abandon it al- 
together. , 

The following work is abridged from a translation of M. 
Bourdon, made by Lieut. Ross, now the distinguished pro- 
fessor of mathematics in Kenyon College, Ohio. 

The Algebra of M. Bourdon, however, has been regarded 
only as a standard or model. The order of arrangement, in 
many parts, has been changed; new rules and new methods 


have been introduced; and all the modifications which have 


4 PREFACE. 


been suggested by teaching and a careful comparison with 
other standard works, have been freely made. It would, per- 
haps, not be just to regard M. Bourdon as responsible for 
the work in its present form. | 

It has been the intention to unite in this work, the scien- 
tific discussions of the French, with the practical methods 
of the English school; that theory and practice, science and 


art, may mutually aid and illustrate each other. 


CHARLES DAVIES. 
WEst Point, June, 1844 


CONTENTS. 


CHAPTER. I. 
PRELIMINARY DEFINITIONS AND REMARKS. 
ARTICLES 
Atcrsra~Definitions—Explanation of the Algebraic Signs......... =, 1—28 
Similar Terms—Reduction of Similar Terms.........-- A Ae Ares aeege, . Coe 
Prublems—Theorems—Definition of—Question.....0...+. Sr eatdss 31—33 


CHAPTER II. 


OF ADDITION, SUBTRACTION, MULTIPLICATION, AND DIVISION. 


Addition—Rule......... A saat Panta cameras ate st hai weed 33—36 
Subtraction—Rule—Remark .....2....+6: ig aura aleiew eal oe PE tere 36—40 
Multiplication—Rule for Monomials ....+...++ Bia aa. On sie ainig ats 40—42 
Rule for Polynomials and Signs ........... Pg ee ee te ea essceee 42—45 
Remarks—Theorems Proved.......see.sceeceecceerece® ears tas 45—48 
Of Factoring Polynomials....... ‘ge prulet Sere PES OM daneece’ | Soe 
Division of Monomials—Rule..... Vere y EES COCR 6 esses 49—52 
Signification of the Symbol a®..............06 Noataily « «ane snc 52—54 
Division of Polynomials—Rule...........-e0+05 Sededdecdledadens, 04-06 
Remarks....... MW Ate sal VitcesseccseecccencsoeEemmmecs) “Oa—OL 


When m is entire, am—b™ is divisible by A—b....00..sesseeeeees 61—62 


CHAPTER III. 


ALGEBRAIC FRACTIONS. 


Definition—Entire Quantity—Mixed Quantity.........sseeesees oe 62—65 


Reduction of Fractions ...........e0+ GLUE ME sok albeit ins aiaa Cote 
To Reduce a Fraction to its Simplest Form....se.scceesccevecces 70 
To Reduc> a Mixed Quantity to a Fraction. ..ces.sseesscereseses 71. 


6 CONTENTS. 


ARTICLES 
To Reduce a Fraction to an entire or Mixed Quantity.......... 72 
To Reduce Fractions to a Common Denominator.....e.++ese0e. 73 
OO ACG TactiOns cnc. ce eee cases Pons wore ieeengne » o.e\ pio» wire 74 
‘Lo. Subtract Fractions. as. sx «siss 5) ste > cc ulstaa ean o tenaa mS 75 
To Multiply Fractions. .....-.seesessecsceccsces Ss alin gee 76 
To Divide Fractions go c5 + «5 cae 6 Cate wie omee ene «ayn oe eon Sa 77 
Results from adding to both Terms of a Fraction......++..seees 78 
CHAPTER IV. 
EQUATIONS OF THE FIRST DEGREE. 

Definition of an Equation—Different Kinds—Properties of Equations 7T9—86 
Principles in the Solution of Equations—Axioms........+e+++- 86—87 
Transformation of Equations—First and Second.......... o asain sip 87—92 
Resolution of Equations of the First Degree—Rule........-+6% i 92—94 
Questions involving Equations of the First Degree........... “xe 94—95 
Equations with two or more Unknown Quantities ..........-+.. 95—96 
Elimination—By Addition—By Subtraction—By Comparison...... 96—103 

Indeterminate Problems—Questions involving two or more unknown 
RPURIITILIOS fo oes Ais Ceo ein at Mini Peer etre - 103—-104 

Theory of Negative Quantities—Explanation of the Terms Nothing 
and Annaity... ¢«. «os a shay ddl n aed e ark» 9m, 0 dup Aigo. eu ae esos  104—114 
Inequalities....... Gia cw Sree Rube rod tenes ree oo Se. o's ale' aie eee 1 


CHAPTER V. 


EXTRACTION OF THE SQUARE ROOT OF NUMBERS.—FORMATION OF 
THE SQUARE AND EXTRACTION OF THE SQUARE ROOT OF ALGE= 
BRAIC QUANTITIES. 


Extraction of the Square Root of Numbers.....s.sseeeesseeeses  116—118 


Of Incommensurable Numbers..........+0-0- - «+ dan oad 118 
Extraction of the Square Root of Fractions.......s.sece-se00s-, I118—124 
Extraction of the Square Root of Algebraic Quantities ....... wits 124 
PUP NADNOUGIGIE ons oe pie ole «' ob. u:0' 9 Se Pare os os ies a cocsecicose . 124-127 
Of Polynomials..... . ep eee ae ge Pe +, fue Sia aidpeineverewie) Fol aiebeeleg 
Caleulus of Radicals of the Second Degree ...........-..ee0: » 130—132 
Addition and Subtraction—Of Radicals.........0.seeeeeeeee-ee. | 132—133 
Multiplication, Division, and Transformation,..... Re es -.  133—137 
CHAPTER bas ' 
EQUATIONS OF THE SECON DEGREE. 
ftions of the Second Degree .........20-.- sis ee 9:0 ae Te bs a | 
Involving two Terms.........- e Shistiniaiey dase aia er - -- 139—140 


Complete Equations of the Second Degree pe: ncvrdey ee | Laieomiad 


CONTENTS. 


Discussion of Equations of the Second Degree....... wa aer es “th: 
Problem of the Lights....... eS oe PCa E TT eee eae ee tte 
re FINO Pp wmationss saad IS 2 US a 
‘Extraction of the Square Root of the Binomial ey om NS 
Equations with two or more Unknown Quantities...........000. 


CHAPTER VII. 


OF PROPORTIONS AND PROGRESSIONS. 


How Quantities may be compared together........ Ve th ee 
Arithmetical Proportion Defined......... ie. 8 aha rane ate See nee ee aa 
(eormenica: Froporuon Defined... i. 0... ec cce cca vee Sierra's anne 
Arithmetical Proportion—Sum of Extremes..... My Pe Raise Sate a 
Arithmetical Progression—Increasing and Decreasing........-... 
Value of Last Term—How to find it............eeeeeee. Cee 
How to find last term in a Decreasing Series .......... 00006 Yds 
Sum of two Terms equi-distant from Extremes .........---s.e0. 
ree Nee “LOTMS., . cient ew hu nccescbetnne 
General Formulas........... Os SHOR eae Baas see Beg ee apie te 
pmPaI TERNS aot DOKI f.% 6 47.'.%.'.''ss' ee eee ew eee ae OW ate aercie se 
To find the Common Difference.........0.......6. Bly Si Seige te 
To find any Number of Means between two Numbers.......... 
The Whole makes a continued Series ........ceccceeecces tel 


GEOMETRICAL PROPORTION , 


AMES CS Sa ore xs PORT 3.05 PIE ES E, 
Proportion Defined 2.0:.'35.4.4.. ee « SS Sia ER SUNT, 
Antecedents and Consequents Defined......-cessccsceesssosecs 
Mean Proportional Defined........... errr GAL UHL Yt, sare 
Proportion by Inversion—or Inversely...... a sr 2 ers ees, 
Proportion by Alternation..... donne de a 2 RUE os de BWLD ING ‘ 
Proportion by Composition..... acaabe sitet ts ¥stat a stanide aoe A vot 
POP OTUGIa DIVISION 06 oe ee PISS Ls we A A IES 
Hqui-multiples are Proportional..... ee Ce ceed bowateu. gen 
Reciprocal Proportion Defined...... baeAMe LSet, SO Me, SR, 19 
Product of Extremes equal to the Product of the Means......... 
To make a Proportion from four Quantities........ SETA AME (2 
Square of middle Term equal to Product of Extremes.......... . 
Four Proportionals are in Proportion by Alternation...... JAMES 
Proportion by Equality of Ratios IRI a os sth ohotonent eles ‘ 
Four Proportionals are Proportional Inversely...... L Ss US 


Four Proportionals are in Proportion by Composition or Division 
Equi-multiples have the sarne Ratio as the Quantities........... 
Proportion by augmenting Antecedent and Consequent..........- 


7 
ARTICLES 
141—150 
150—151 
151—154 
154—157 
157—159 


8 CONTENTS. . 


ARTICLES 
Proportion by adding Antecedent and Consequent........esseees 187 
The Powers of Proportionals are in Proportion.......cceessecees 188 
The Products of Proportionals are in Proportion........s.seeees 189 
Geometrical Progression Defined .............-6- <bean aaee 196 
Value of the Last Term........... ITT Ry er nt 191 
To find the Sum of the Series......... Were 192 
To find the Sum of the Terms of a Decreasing Progression...... 193 
When the Sum becomes 2... . $00" SPA Rie coe 194 
To find any Number of mean Proportionals between two Num- 
DEM As seta ke cee ee Sees Cece cet cces ods sak eatin ls meee See 195 
Progressions having an Infinite Number of Terms...... do eu epte 196 


CHAPTER VIII. 


FORMATION OF POWERS, AND EXTRACTION OF ROOTS OF ANY DE- 
GREE WHATEVER.—CALCULUS OF RADICALS.——-INDETERMINATE CO- 
EFFICIENTS. 


Hormation: of SPOWECIS. js/s0.0.5 00.000 «04 4.00 (0.0 wine v/s ee sees.  197—199 
Theory of Permutations and Combinations............. view edeat 4190-202 
Binomial Theorem ...... © ca pun meine «e's serie os <ienn 202—206 
Consequences of Binomial Theorem...........se0.cegeceecces ‘* 206—209 
Extraction of the Cube Roots of Numbers............secseeees 209—213 
To Extract the nt Root of a Whole Number.................. 213—215 
Extraction of Roots by. Approximation: .: %...'.% «.< » auenisn ener 215—218 
Cube Root of Decimal Fractions. ....2 .. .. + +0 bse kode 6c 218 
Any Root. of .a,.Decimal, Fraction. «+: «.«6s «0145 sealer 219 
Formation of Powers and Extraction of Roots of Algebraic Quan- 

THERES «oo ne os 00 oyna oe 0's 0 9.00 0 elem ‘oo alte ie letteh One tn 220 
Of Monomials—-Of Polynomials. ......0.2s0c,cces me muuls une ges 221—224 
Calculus of Radicals—Transformation of Radicals..... en ee 224— 227 
Addition and Subtraction of Radicals.......2 .s.5ia, see eee 227 
Multiplication and Division ...».% 20+... 0s «ssi een 228 
Formation of Powers and Extraction of Roots.......... sey ies 229 
Different Roots. of Unity .....%..26.e. saves «> eee SVCk Cte. 230—232 
Modifications of the Rules for Radicals...........200.ssceesees 232 
REO Of LXPON ENS, « «ccikie:. sdk gestae Ae C.0. Re 233 
Multiplication of Quantities with any Exponent.............. as 234 
PRIMA ks fp oe oe ae ee oven phe koa ai lye’ > stale SRR sa alain antag 235 
Rprmation of Powerg...........<atleeee saya tae bbtChiafaee aS 5 ote + aa 236 
PPetrAchene ar Oth oe eas ®. cha lelas «sie 4 e0S 237 
Method of Indeterminate Co-efficients.........sssseccsescnsees 238—24? 
Recurring Series..... sh oe cams } cab ane sa i dae 243 
Binomial Theorem for any Exponent..........+ese0- o a eget <i 244—245 


Applications of the Binomial Theorem......... Peer foe 


‘ CONTENTS. 
Summation of Series..... ne che wae cass ocean. wdaRee fh 
Summation of Infinite Series...........0¢% tinea wierd ewe owas Be 


CHAPTER IX. 


9 
ARTICLES 
246 

247 


CONTINUED FRACTIONS.—EXPONENTIAL QUANTITIES.—LOGARITHMS. 


—FORMULAS FOR INTEREST. 


Continued Fractions...... fe Peer eee re Cre a nied waices 
Exponential Quantities..........4. Rinse nes aon ass da tew calves 
Theory of Logarithms......... as elaine few hi sinie Rigeletes 6 Sa dnwinte s 
MeO epiea tO ANG DIVISION... 0 accccecicndecenease aig cae meets 
Formation of Powers and Extraction of Roots..........seeesee. 
re MM CNRABO TE re. Le VEL a IAels & cicle.als disilss suds is.cc dees 
Logarithmic and Exponential Series—Modulus........0....0+ee 
Ea se AMIE ETI OR ho. iia «<a cieiele'dis, « an:at bis si¥ te vinsidided oo 
MRC RII MINI Said's uvns Va oie 6 occ ecncctes deve Se Bees ae see 
Off lntsrest cosets «\< 6 es ee te OE Se oe pa ctelewearente 
4 CHAPTER X. 

GENERAL THEORY OF EQUATIONS. 
peneral) Properties Of Fquations. ....ccoccceccscacessccmscsese 
Of the Greatest Common Divisor...... aie wate cies ike deaueie 
Transformation of Equations.........0-ses.ee. gic waa slaw eleths © ; 
Remarks on Transformations . ..........seee0s Sinide wee saele 
Derived Polynomials............. Be estaba es Fick SA occ cceee 
PIT TLOUUS Suc a vccacctssses aaete SER cece ee ewes os ce eee ‘ 
PO UIONN OM as cece esc cdecese o aPe tad bcy orate. aieners cctoeeus te 
By Means of Indeterminate Multipliers ......c...ccesccescees 
By Means of the Common Divisor...... = ave Ninrmie esse S ve cannnegre 
Method of finding the Final Equation....... eiap lis <.nide We. eustes ale 


CHAPTER XI. 


248—254 

255 
256—258 
258—260 
260—262 
262—266 
266—270 
270—272 
. 272 

273 


274—285 
285—294 
294—296 

296 
297—300 
300—303 

303 

304 
305—307 
307—309 


RESOLUTION OF NUMERICAL EQUATIONS.—STURMS’ THEOREM. 


Fust Principle. ..0ss. iss | Batol selec States ea bds Whenbiasie waite ont 
BOG  ETINCIPIC:. coed cbaiwiew sc cee oe attiaiteds gc GRR west eine Stele oe 
moird, Principle . ....01.a0s PUN lats Whieis ivi wicca higtalaca a eum Te 
Dmmre Of teal Roots: ...v 0. cus sc cceeccs bs tives eae netin week see's 


Ordinary Limits of Positive Roots ......e..ee0. 


eeooeveeneseese eee 


312 
314—317 
317 


19 CONTENTS. 


Smallest Limit in Entire Numbers..... SdedscclPPoes aa Sawer 
Superior Limit of Negative Roots—Inferior Limit of Positive and 

(Negative Roots.....csccecs-cccvsceses Sere rr 
CLONSOGIEN COS e's isin le aisle sia ieisiess'e's arse cin 00 vine wiwiple ates mE 0+ kee 
Descartes’ Rule......... Py Cry ee ee dele Jee vies ews SG Oe mM 
Commensurable Roots of Numerical Equations.......... etsy eee a ° 
Sturms’) Theorem. @sdg sss seks AA A see dtl PLS ak 
Young’s Method of resolving Cubic Equations...... Miweneweids b65% 
Method of Resolving Higher Equations........... sia Popecee 

é 


ARTICLES 
318 


319 
320—327 
327—330 
330—333 


 333—341 


342—345 
345 


. ELEMENTS 


OF 


ie. 98 Mon [bos (FoAH oo Brow Bee sb, 


CHAPTER I. 
PRELIMINARY DEFINITIONS AND REMARKS. 


1. QuANTITY is a general’ term applied to everything which 
can be estimated or measured. 


2. Maruematics is the science which treats of the properties 
and relations of quantities. 


3. ALGEBRA is that branch of mathematics in which the quanti- 
ties considered are represented by letters, and the operations to 
be performed upon them are indicated by signs. ‘The letters and 
signs are called symbols. 


4. The sign +, is called plus, and when placed between two 
quantities indicates that they are to be added together. ‘Thus, 
§ + 5 is read, 9 plus 5, and indicates that the quantity repre- 
sented by 5 is to be added to the quantity represented by 9. 

In like manner, a+ 0 is read, a plus 0b, and denotes that the 
quantity represented by d is to be added to the quantity repre- 
sented by a. 

5. The sign —, is called minus, and indicates that one quantity 
is to be subtracted from another. Thus, 9 — 5 is read, 9 minus 5 
or 9 diminished by 5. 

In like manner, a — } is read, a minus 6, or a diminished by } 

6: The sign xX, is calla the sign of multiplication. When 
placed between two quantities, it denotes that they are to be mul- 
tiplied together. Thus, 36 x 25, denotes that 36 is to be multi- 
olied by 25. The multiplication of two quantities may also be 

* 


12 ELEMENTS, OF ALGEBRA. [CHAP. 1. 


indicated by placing a point between them. Thus, 36.25 is the 
same as 36 X 25, and is read, 36 multiplied by 25, or the prod- 
uct of 36 by 25. 

7. The multiplication of quantities, which are represented by 
letters, is generally indicated by simply writing the letters one 
after the other, without interposing any sign. ‘Thus, ; 


ab isthe same as a X Bb, oras ab; 
and abc, the same as a x b xc, or as a.b.c. 

It is plain that the notation ab, or abc, cannot be employed 
when the quantities are represented by figures. For, if it were re- 
quired to express the product of 5 by 6, we could not write 5 6, 
without confounding the product with the number 56. 


8. In the product of several letters, as abc, the single letters, 
a, 6, and c, are called factors. ‘Thus, in the product ad, there 
are two factors, a and 6; in the product acd, there are three,. 
a, c, and d. 

9. There are three signs used to denote division. Thus, 

a — 6 denotes that a is to be divided by 6. 
= denotes that a is to be divided by 4. 
a\b denotes that a is to be divided by 5. 


10. The sign =, is called the sign of equality, and is read, us 
equal to. When placed between two quantities, it denotes that 
they are equal to each other. Thus, 9 —5=4: that is, 9 mi- 
nus 5 is equal to 4: Also, a+ 6=c, indicates that the sum of, 
the quantities represented by a and 4, is equal to the quantity de- 
noted by ec. 

11. The sign >, is called the sign of inequality, and is used 
to express that one quantity is greater or less than another. 

Thus, a> 6 is read, a greater than 0; and a< 6 is read, 
a less than 6; that is, the opening of the sign is turned toward 
the greater quantity. 


12. If a quantity is added to itself several times, as 


ata+a+@+ a, 


it is generally written but once, and a figure is then placed before 
it, to show how many times it is taken. Thus, 


at+a+ata+a= 5a. 


CHAP. I.] DEFINITIONS AND REMARKS. rg 


The number 5 is called the co-efficient of a, and denotes that a is 
taken 5 times. 

Hence, a co-efficient is a number prefixed to a quantity, denoting 
the number of times which the quantity is taken. The co-efficient 
also indicates the number of times plus one, that the quantity is 
added to itself. When no co-efficient is written, the co-efficient 
1 is always understood. Thus, a = la. 


13. If a quantity be multiplied continually by itself, as 
VRS hy al Me Os Se ae 


the product is generally expressed by writing the letter once, and 
placing a number to the right of, and a little above it: thus, 


axaxaxaxa-—a. 


The number 5 is called the exponent of a, and denotes the 
number of times which a enters into the product as a factor. 

Hence, the exponent of a quantity shows how many times the 
quantity is a factor. It also indicates the number of times, plus 
one, that the quantity is to be multiplied by itself. When no ex- 
ponent is written, the exponent 1 is always understood. 


14. The product resulting from the multiplication of a quantity 
‘by itself any number of times, is called the power of that quantity: 
and the exponent denotes the degree of the power. For example, 


a1 — a is the first power of a, 

a* = a X a is the seeond power, or square of a, 
a? — aX a Xa is the third power, or cube of a, 
a*—axXaXa Xa is the fourth power of a, 
a@—axaxaxXaxXa is the fifth power of a, 


in which the exponents of the powers are, 1, 2, 3, 4, and 5; and 
the powers themselves, are the results of the multiplications. It 
should be observed that the exponent of a power is always greater 
by unity than the number of multiplications. 


15. As an example of the use of the exponent in algebra, let 
it be required to express that a number a is to be multiplied 
three times by itself, that this product is then to be multiplied 
three times by 3), and this new product’ twice by ¢; we should 
write | 

gia! x) Oe PD EK bok exe at abFe?: 


If it were further required to repeat this result a certain num- 


14 LLEMEN?TS OF ALGEBRA. - (CHAP. I. 


ber of umes, say seven times, that is, to add it to itself six times, 
we should simply write 
Tate 

This example shows the brevity of the algebraic language. 

16. The root of a quantity, is a quantity which being multi- 
plied by itself a certain number of times, will produce the given 
quantity. 

The sign / , is called the radical sign, and when prefixed 
to a quantity, indicates that its root is to be extracted. Thus, 


‘/ a or simply +f a denotes the square root of a. 
V a denotes the cube root of a. 
a/ a denotes the fourth root of a. 


The number placed over the radical sign is called the index 
of the root. Thus, 2 is the index of the square root, 3 of the’ 
cube root, 4 of the fourth root, &c. 


17. The reciprocal of a quantity, is unity divided by that quan: 
tity. ‘Thus, 


the reciprocal of a; 


and, 





+}e a[e 


aa) is the reciprocal of a+ 8. 


18. Every quantity written in algebraic language, that is, with 
the aid of letters and signs, is called an algebraic quantity, or thie 
alyehrawr expression of a quantity. Thus, . 


94 ; is the algebraic expression of three times the 
quantity denoted by a; 
; is the algebraic expression of five times the 
square of a; 
is the algebraic expression of seven times the 
0} product of the cube of a by the square of b; 
is the algebraic expression of the difference 
between three times a and five times 6; 
is the algebraic eXpression of twice the square 
of a, diminished by three times the product 
of a by b, augmented by four times the square 
of b. 


Ya? — 3ab + 40? 


CHAP. I.] DEFINITIONS AND REMARKS. 1 


ao 


19. -A single algebraic expression, not connected with any other 
by the sign of addition or subtraction, is called a monomial, or 
simply, a term. 

Thus, 3a, 5a’, 7a’b?, are monomials, or single terms. 


20. An algebraic expression composed of two or more terms, 
separated by the sign + or —, is called a polynomial. 

For ‘example, 3a — 54, and 2a? — 3cb + 407, are polynomials. 

A polynomial composed of two terms, is called a binomial; and 
a polynomial of three terms is called a trinomuial. 


21. The numerical value of an algebraic expression, is the num- 
ber obtained by giving a particular value to each letter which en- 
ters it, and performing the arithmetical operations indicated. This 
numerical value will depend on the particular values attributed to 
the letters, and will generally vary with them. 

For example, the numerical value of 2a°, will be 54 if we make 
Cee er ee, and 2.x. 27 i= 54, ; 

The numerical value of the same expression is 250 when we 
Mpkenee Oe tor, Oo = 125, and 2 x 129'=-250. 


22. It has been said, that the numerical value of an algebraic 
expression generally varies with the values of the letters which 
enter it: it does not, however, always do so. Thus, in the ex- 

. pression a — J, so long as a and & are increased or diminished 


by the ‘same number, the value of the expression will not he 
changed. 


For example, make a= 7 and b = 4: there results a — b = 3. 
Now, make a=7+5=12, and b6=4+5=9, and there 
results, as before, a — 6 = 12 —9 =3. 


23. Of the different terms which compose a polynomial, some 
are preceded by the sign +, and others by the sign —. The 
first are called additive terms, the others, subtractive terms. 

When the first term of a polynomial is plus, the sign is gener- 
ally omitted; and when no sign is written, it is always under- 
stood to be affected by the sign +. 


24. The numerical. value of a polynomial is not affected by 
changing the order of its terms, provided the signs of all the, 
terms remain unchanged. For example, the polynomial 

403 — 30% + 5ac? = Sac? ~ Balt Ac? = - 30% + Sac? + 40 


16 ELEMENTS OF ALGEBRA. [CHAP. I. 


25. Each of the literal factors which compose a term, is called“ 
a dimension of the term; and the degree of a term is the number 
of these factors or dimensions. Thus, 

3a is a term of one dimension, or of the first degree. 

5ab is’a term of two dimensions, or of the second degree. 

7a%bc? = Taaabce is of six dimensions, or of the sixth degree. 

In general, the degree, or the number of dimensions of a term, ts 
determined by taking the sum. of the exponents of the letters which 
enter this term. For example, the term 8a%bed* is of the seventh 
degree, since the sum of the exponents, 2+1+1+3=7. 

26. A polynomial is said to be homogeneous, when all its terms 
are of the same degree. The polynomial 


3a — 2b+ is of the first degree and homogeneous. 
— dab + B is of the second degree and homogeneous. 
5a’c — 4c? + 2c*d is of the third degree and homogeneous. . 
8a3 —4ab+c is not homogeneous. — 


27. A vinculum or ‘bar 





, or a parenthesis ( ), is used to 
express that all the terms of a polynomial are to be considered 
together. Thus, a+ b+exb, or (a+6+c) xb denotes 
that the trinomial a+ 6+ is to be multiplied by 4; also, 
a+b+texc+td+f or (a+tb+c) x (ec+d+f) 
denotes that the trinomial @+ 6+ c¢ is to be multiplied by the 
trinomial ¢ + d+ f. 
When the parenthesis is used, the sign of multiplication is usually 
omitted. Thus, (a+65-+c) x 6 is the same as (a+b+c))d 
The bar is also sometimes placed vertically. Thus, 
+ a\ ex Pe ae 
+6) is the same as (2 +6+c)a, or, a+b+cxa@. - 
+e 
28. Those terms of a polynomial which are composed of the 
same letters, affected with the same exponents, are called similar 
terms. | 
Thus, in the polynomial 
Tab + 3ab — 4a3b? + 52383, 
the terms 7ab and 3ab, are similar, and so also are the terms 
_— 4a°b? and 5a%b?, the letters and exponerts in each being the 
same. But in the binomial 
8a2b + 7ab?, 








' 


CHAP. I.] DEFINITIONS AND REMARKS. 17 


7 


the terms are not similar; for, although they are composed of the 
same letters, yet the same letters are not affected with the same 
exponents. 

29. When a polynomial contains several similar terms, it may 
often be reduced to a simpler form. 

Take the polynomial 4a°b — 3a’c + Ta’e — 20d. 

It may be written (Art. 24) 4a7b — 2a*b + Ta’e — 3a°e. 

But. 4a2b — 2a7b reduces to 2a7b,, and 7Va’c — 3a%c to 4a%c. 

Hence, 4a2b — 3a’c + Ta*c — 2a*b = 2a*b + 4a?e. 

When we have a polynomial having similar terms, as 

+ 2a%bc? — 4a%bc? + 6arbc? — 8a%bc? + 1larbec?, 

unite the additive and subtractive terms separately: thus, 


Additive terms. Subtractive terms. 
, bt 2a%bc? — 4arbe? 
+ 6a%bc? — B8a%bc? 
++ lla%bc? fin 12a3bc2 
+ 19a%bc? 


Hence, the given polynomial reduces to 
19a%bc? — 12a%be* = Tarbe?. 
lt may happen that the co-efficient of the subtractive term, ob- 
tained as above, will exceed that of the additive term. In that 
case, subtract the positive co-efficient from the negative, prefia the 
minus sign to the remainder, and then annex the literal part. 


In the polynomial 3a7b + 2a*b — 5a*b — 3a%b 
+ 3ab — 5a*b 
+ 2atb — 3a%b 
+ 5a2b — 8a*b 
But, — 8a*b = — 5a2b — 3a2b: hence 
5a?b— 8u2b = 5a2b — 5a2b — 3a2b = — 3a2b. 


_ Hence, for the reduction of the similar terms of a polynomial, 
we have the following 
RULE. 

I. Add together the co-efficients of ull the additive terms, and an- 
nex lo ther sum the literal part: form a single subtractive term in 
the same manner. 

Il. Then, subtract the less co-efficient from the greater, and to the 
remainder prefix the sign of the greater co-efficient, and annex the 
literal part. i 


aP 
a. 
' a } 


18 ELEMENTS OF ALGEBRA. (CHAP. 1 


EXAMPLES. 


1. Reduce the polynomial 4a*b — 8a*b — 9ab + 11lab to its 


simplest form. Ans. — 2a*b. 
2. Reduce the polynomial abc? — abc? — Tabc* — 8abc? + Gabe? 
to its simplest form. Ans. — 3abc?. 
3. Reduce the polynomial 9cb? — 8ac? + 15cb? + 8ca + Quc? 
— 24cb? to its simplest form. Ans. ac* + 8ca. 
4. Reduce the polynomial 6ac? — 5ah3 + Tac? — 3ab3 — 13ac? 
+ 18ab? to its simplest form. Ans. 10ab3. 
5. Reduce the polynomial abc? — abc + 5ac? — 9Yabc? + Gabe 
— 8ac? to its simplest form. Ans, — 8abe? + 5abe —3ac?. 


Remarx.—lIt should be observed that the reduction affects only 
the co-efficients, and not the exponents. 

The reduction of similar terms is an operation peculiar to al- 
gebra. Such reductions are constantly made in Algebraic Addition, 
Subtraction, Multiplication, and Division. 


30. In the operations of algebra, there are two kinds of quan- 
tities which must be distinguished from each other, viz. 

Ist, Those whose values are’ known or given, and which are 
called known quantities ; and 

2dly, Those whose values are unknown, which are called un- 
known quantities. 

The known quantities are represented by the first letters of 
the alphabet, a, 6, c, d, &c.; and the unknown, by the final 
letters, x, y, 2, &c. 


31. A problem is a question proposed which requires a solution. 
It is said to be solved when the values of the quantities sought 
are discovered or found. 


A theorem is a general truth, which is proved by a course of 
reasoning called a demonstration. 


32. The following question will tend to show the utility of the 
algebraic analysis. 
Question. 


The sum of two numbers is 67, and their difference 19; what are 
the two numbers? 





CHAP. 1.] DEFINITIONS AND REMARKS. 19 


Solution. 

Let us first establish, by the aid of the algebraic symbols, the 
connexion which exists between the given and unknown num- 
bers of the question. 

If the least of the two numbers were known, the greater could 
be found by adding to it the difference 19; or in other words, 
the less number, plus 19, is equal to the greater. 

If, then, we make «2 = the less number, 

x-+19= the greater, 
and . 2n +19 = the sum. 

But from the enunciation, this sum is to be equal to 67. There- 
fore we have 

g04- 19 =! 67: 

Now, if 2x augmented by 19, gives 67, 2x alone is equal to 

67 minus 19, or 
| 2x = 67 — 19, or performing the subtraction, 24 = 48. 
Hence, x is equal to half of 48, that is, 


48 
adeeb gy Dh 
ices 


The least number being 24, the greater is 
e+19= 24+ 19 = 43. 
And, indeed, we have 
43 +24 = 67, and 43 —24=—19. 


Another Solution. 


Let x represent the greater number ; 

then, x — 19 will represent the less 

and Qo — 19 = 67; whence, 2x = 67+ 19 = 86; 
86 

therefore, t=> = 43 — the greater, 


and consequently, #«—19=43 —19=24= the less 


General Solution of this Problem. 


The sum of two numbers is a, and their difference is 6. What 
are the two numbers ? 
Let x = the less number ; 
then will, x-+ == the greater. 


20° ELEMENTS OF ALGEBRA. 7 (CHAP, I. 


Then, by the conditions of the question 
22 + 6 =a, the sum of the numbers ; 


ab we b 

2) Benet 

And by adding 0 to each side of the equality, we obtain the 
greater number, 





therefore, 22 =a—b and x= 


a a b 
be a 
fir 2 Sane 
Hence we have 
Ee > +- _ = the greater number, 
a b 
and Fy ions ae — > = the less. 


As the form of these results is independent of any particular 
values attributed to the letters a and b, the expressions are called 
formulas, and may be regarded as comprehending the solution of 
all questions of the same nature, differing only in the numerical 
values of the given quantities. Hence, 

A formula is the algebraic enunciation of a general rule, or 
principle. 

The principles enunciated by the formulas above, are these: 

The greater of any two numbers is equal to half their sum in- 
ereased by half their difference; and the less, is equal to half their 
sum diminished by half their difference. 

To apply these formulas to the case in which the sum is 237 
and difference 99, we have 


237 99 237+99 336 


the greater number = Vitinm=cis 168 ; 
231 99° 237 — UO ee 
and the less = gig =" ee 69; 


and these are the true numbers; for, 
168 + 69 = 237 which is the given sum, 

and 168 —69= 99 which is the given difference. 

From the preceding explanations, we see that Algebra is a 
language composed of a series of symbols, by the aid of which, 
we can abridge and generalize the operations required in the solu- 
tion of problems, and the reasonings pursued in the demonstration 
of theorems 


a 


CHA] 1.) ADDITION: 21 


CHAPTER II. 


OF ADDITION, SUBTRACTION, MULTIPLICATION, AND DIVISION. 


ADDITION. 


33. AppiITION, in algebra, consists in finding the simplest equiv- 
alent expression for several algebraic quantities. Such equivalent 
expression is called their swm. 





34. Let it be required to add together ise 
5b 
the monomials, 
2c 
The result of the addition is - - - 34+ 5b + 2¢ 
an expression which cannot be reduced to a more simple form. 
, 4a*b3 
Again, add together the monomials - 2a?b3 
7a?hs 
The result, after reducing (Art. 29), is - 132253 
3a? — 4ab 


Let it be required to find the sum of 


the expressions, 2a? — 3ab + BD? 
2ab — 5b? 
Their sum, after reducing (Art. 29),is - 5a? — 5ab — 40? 


35. As a course of reasoning similar to the above would apply 
to all algebraic expressions, we deduce, for the addition of alge- 
braic quantities, the following general | 


RULE. 


[. Write down the quantities to be added, with their respective 
signs, so that the similar terms shall fall under each other. 

Il. Reduce the sumilar terms, and annex to the results those terms 
which cannot be reduced, giving to each term its respective sign. 


a ELEMENTS OF ALGEBRA. [CHAP. II. 


EXAMPLES. 


1. Add together the polynomials, 
3a2 — 2b2 — 4ab, 5a? — 6? + 2ab, and 3ab — 3c? — 207. 


The term 3a? being similar to 5a’, ao Ah oe ie 
we write 8a? for the result of the re- 5a? + Qah— ¥ 
duction of these two terms, at the same 4. 3ab — 282 — 3¢2 











time slightly crossing them as in the ere Oe Bake REC 
terms of the example. Tat a ee eae bo 

Passing then to the term — 4ab, which is similar to the two 
terms + 2ab and + 3ab, the three reduce to + ab, which is 
placed after 8a?, and the terms crossed like the first term. Pas 
sing then to the terms involving 57, we find their sum to be — 50’, 
after which we write — 3c?. , 

The marks are drawn across the terms, that none of them 
may be overlooked and omitted. 




















(2). (3). 
7x + 3ab+ 2¢ 16a2b? + be — 2abe 
» — 32 — 3ab— 5e — 4a?b? — 9be + Gabe 
5a — 9ab — Ye — 9a2h? + be+ abe 
Sum. 92x — 9ab — 12c _3a?b? — The + 5abe 
(4). (5). 
a+ ab— cd+ f 6ab+ cd+d 
3a + 5ab — 6ced — ff 3ab+ Sed+y 
— 5a — 6ab + bed — 7f —4ab+ 6cd+ x 
— at ab+ cd+ 4f — 5ab — 12cd — y 
ahaa ae. 0 2 5F, 0 On abun crud 





6. Add together 34+ 6, 3a+ 3b, —9a— 7b, 6a+96, and 
8a + 3) + 8c. Ans. lla + 9b + 8c. 


7. Add together 3ax + 3ac+ f, —9ax+ 7a+d, + 6ax + 3ae 
+ 3f, 8ax + 13ac + Of, and — 14f + 3aa. 
Ans. llax + 19ac —f+ 7a + d. 
8. Add together the polynomials, 3a%c + 5ab, Ta%e — 3ab + 3ac, 
5a’c — 6ah + Yac, and — 8a’c + ab—12ac. Ans. Vac — 3ab. 
9. Add the polynomials 19a?x°b — 12a3cb, 5a?a3b + 15a3cb 
— 10ax, — 2a?x*h — 13a%ch, and — 18a2x3b — 12a3cb + Qaz. 
Ans. 4a?x°b — 22a°ch — az 





CHAP. II.] ADDITION. 23 


10. Add together 3a + b+ ¢, 5a +4 26+ 3ac, a+c-+ ac, and 


P— 34 — 9ac — 8b. Ans. 6a — 5b + 2c — 5ae. 
11. Add together 5a2b + 6cx + 9bc?, 7cx — 8a*b, and — ldcx 

— Q9bc? +- 207d. Ans. — a*b — 2cx. 
12. Add together 8ax + 5ab + 3a2b2c?, — 18ax + 6a? + 10ab, 

and 10ax — 15ab — 6a?6?c?. Ans. — 3a?b2c? + 6a?. 
13. What is the sum of 41a%b?c — 27abe —14a’y and 10a6e 

+ 9abe? Ans. 51a%b?c — 18abe — 14a*y. 
14. What is the sum of 18abe — 9ab + 6c? — 3c + 9ax and 

9abe' + 3c — 9ax? Ans. 27abe — 9ab + 6c. 
15. What is the sum of 8abe + b3a —2cx —6xy and Tex 

— ry — 1363a! Ans. 8abe — 12b?a + Sex —Taxy. 


16. What is the sum of 9a%e — l4aby + 15076? and — a 
— 8a7b?? Ans. 8a*c — l4aby + 7a?b?. 
17. What is the sum of 17a'b? + 9a°b — 3a?, — 1405}? + Ya? 
— 9a3, — 15a°b + 7a°b? — a3, and 14a3b — 19a3h? 
Ans, ——————_. 
18. What is the sum of 3ax?— 9ax? — 17ary, — Gaz? + 18ax3 
+ 34axy, and 7a°d + 3ax3 — Tax? + 4bcx? Ans. 
19. Add together 3a? + 5a?b?c? — 9a3x, 7a? — 8ab*c? — 10a3z, 
10ab + 16a7b2c? + 19a3x. Ans. 10a? + 13a2b2c? + 10ad. 
20. Add together 7a2b — 3abe — 8b?e'— 9c3 + cd*, 8abe — 5a%b 
+ 3c3 — 4b2c + cd?, and 4a2b — 8c3 + 9b%c — 3d?. 
Ans. 6a?b + 5abe — 3b2c — 14c3 + 2cd? — 3d3. 
21. Add together — 18a°b + 2ab+ + 6a7b? — 8ab* +- 7a3b — 5ab? 
— 5a°b + 6ab* + 114782. Ans. — 16a°b + 12a2}?., 
22. What is the sum of 3a%b’c — 16atx — 9ax*d, + 6a%be 
— 6axr'd + 17atx, + 16ax°d — atx — 8ab?c? 
Ans. a3b?c + ax3d. 
23. What is the sum of the following terms: viz., 8a° — 10a*b 
— 16a°b? + 40263 — 1204) + 1503b2 + 240253 — abs — 16a°b? 
of 20a2b3 + 32ab+ — 855? 
Ans. 8a5 — 22ath — 17u3b? + 480753 + 26ab* — 805, 


24 ELEMENTS OF ALGEBRA. {CHAP. Il. 


SUBTRACTION: 


36. Susrraction, in Algebra, consists in finding the simplest 
expression for the difference between two algebraic quantities. 

Let it be required to subtract 4b from 5a. MHere,.as the quan- 
tities are not similar, their difference can only be imdicated, and 
we write 

5a — 4b. 

Again, let it be required to subtract 4a°b from 7ab. These 
rms being similar, one of them may be taken from the other, 
and their true difference is expressed by 

7a2b — 4a°b = 3a%b. 


For another example, from - - - 4a 
take the binomial - - - - 2b — 3c. 
The subtraction may be indicated thus, - 4a — (2h — 3c); 


that is, the quantity to be subtracted may be | 4, 95 4 3¢ » 
placed within a parenthesis, and written after the | True remainder. 
other quantity, with a minus sign. 

Now, in order to express this difference by a single polyno- 
mial, let us see what the nature of the question requires. 

From 4a, we are to subtract the difference between 24 and 3c, 
and if 4 and ¢ were given numerically, that difference would be 
known; but since 3c cannot be taken from 20, 24 is first Subtracted 
from 4a, which gives 4a — 2b. Now, in subtracting the number 
of units contained in 2d, the number taken away from 4a, is too 
great by the number of units contained in 3c, and the result 
4a — 2b is therefore too small by 3c; this remainder must there- 
fore be corrected by adding 3c to it. Hence, there results from 
the proposed subtraction 4a — 2b + 3c. 


37. Hence, for the subtraction of algebraic quantities, we have 
the following general | 


RULE. 


I. Write the quantity to be subtracted under that from which tt 
is to be taken, placing the similar terms, tf there are any, under 
each other. 

Il. Change the signs of all the terms of the polynomial to be 
subtracted, or conceive them to be changed, and then reduce the poly 
nomial result to its simplest form 


CHAP. II.] SUBTRACTION. 25 


EXAMPLES. 
(1). a24 (1). 
From - 6ac — 5ab+ c? ay 6ac — 5ab + c? 
Take - 3ac + 3ab — 7c ee — 3ac — 3ab + 7c 
Remainder 3ac — 8ab + c? 4+ Te. bet 3ac — 8ab + 2 + Te 


(2). (3). 























From - 1602 — 5bc + Tac 19abe — 16ax — Saxy 

Take - 14a2-+ 5bc + 8ac 17abe + Tax — 15axy 

Remainder 2a? —10be — ac 2abe — 23ax + 10acy 
(4). (5). 

From - 5a? — 4a2b+ 302c 4ab— cd-+ 3a? 

Take - —2a?-+- 3a2b — 80b2c 5ah — 4ed + 3a? + 5b? 

Remainder 723 — 702) + 11b%c — “ab + 3cd + 0 — 5b?. 


6. From 3a2x — 13abe + 7a?, take Q9a2x — 13abe. 
Ans. — 6a*x + 7a?. 
7. From 51a%b?e — 18abe — 14a?y, take 41a%b2c — 27abe 
— 14a*y. | Ans. 10a%b?c + Qabe. 


8. From 27abe — 9ab + 6c?, take QYabe + 3c — Yaz. 
Ans. 18abe — 9ab + 6c? — 3c + 9az. 


9. From 8abe — 12b3a + Seca — Try, take Tex — xy — 13D8a. 
Ans. 8abe + b?a — 2cx — 6xy. 


10. From 8a?c — l4aby + 7ab?, take 9a’e — 14aby + 15a?b?. 


Ans. — a’c — 8a?b?2. 
11. From 9aSx? — 13 + 20ab3?x — 4b%cx?, take 3%cx? + 9uSx? 

— 6 + 3ab3x. Ans. 17ab3x — 7b®ex? — 7. 
12. From 5a*t — 7a3b? — 3c'd? + 7d, take 3a*t — 3a? — 7c%d? 

_ — 15a3b?. Ans. 2a* + 8a5b? + 4c5d? + 7d + 3a?. 


13. From 51a2b? — 48a35 + 10a*, take 10a* — 8a°b — 6a?b?. 
Ans. 57a2b2 — 40a%b. 
14. From 21a y? + 25x?y3 + 68ry* — 40y5, take 6427y° 
+ 48xy* — 407’. Ans. 20xy* — 39x?y3 + 21x3y?. 
15. From 53a? — 1522y3 — 18aty — 5625, take — 15a?y° 
+ 1833y? + 24aty. Ans. 35a3y? — 42aty — 56x° 





26 ELEMENTS OF ALGEBRA. [CHAP. II. 


38. From what has preceded, we see that polynomials may be 
subjected to certain transformations. 


For example - - 6a? — 3ab + 2b? — 2be, 
may be written - 6a? — (3ab — 2b? + 2be). 
In like manner - - 7a? — 8a2b — 4b%c + 683, 
may be written © - 7a? — (8a7b + 4b%c — 663); 
or, again, - - - 7a* — 8a*b — (4b?c — 65°). 
Also, - - = = 8a? — 6a2b? + 50763, 
becomes - - - 8a? — (6a?b? — 5a?b'). 
Also, -. - - + - 9Qa%c} — 8at + 6? —e, 
may be written - Q9a%c? — (8at — b? + c); 


or, it may be written 9a’c3 + b? — (8at + c). 


These transformations consist in decomposing a polynomial into 
two, parts, separated from each other by the sign —. 

It will be observed that the sign of each term which is placed 
within the parenthesis is changed. Hence, if wé have one or 
more terms included within a parenthesis having the minus sign 
before it, if the parenthesis is omitted, the signs of all the terms 
must be changed. 


Thus, 4a — (6ab — 3c — 2h), 
is equal to 4a — 6ab + 3¢ + 2d. 
Also 6ab — (— 4ac + 3d — 4ab), 
is equal to 6ab + 4ac — 3d + 4ab. 


39. RemarK.—From what has been shown in addition and 
subtraction, we deduce the following principles. 

Ist. In Algebra, the words add and sum do not always, as in 
arithmetic, convey the idea of augmentation. For, a — b, which 
may result from the addition of — 6b to a, is properly speaking, 
the arithmetical difference between the number of units expressed 
by a, and the number of units expressed by 6. Consequently, 
this result is numerically less than a. 

To distinguish this sum from an arithmetical sum, it is called 
the algebraic sum. j 

Thus, the polynomial, 2a? — 3a?b + 367e. 
is an algebraic sum, so long as it is considered as the result, 
of the union of the monomials 2u°, — 3a°b, + 36?c, with their 
respective signs; but, in its proper acceptation, it is the arithmeti- 





CHAP. II.] MULTIPLICATION. ed 


cal difference between the sum of the units contained in the ad- 
ditive terms, and the sum of the units contained in the subtractive 
terms. 

It follows from this, that an algebraic sum may, in the numeri- 
cal applications, be reduced to a negative number, or a number 
affected with the sign —. 

2d. The words subtraction and difference, do not always convey 
the idea of diminution. For, the difference between + a and — b 
being a —(— 6) =a-+4, is numerically greater than a. This 
result 1s an algebraic difference. 


MULTIPLICATION. 


49. AuGEBRAIC multiplication has the same object as arithmeti- 
cal, viz., to repeat the multiplicand as many times as there are 
units in the multipher. The multiplicand and multiplier are called 
factors. 

It is proved in Arithmetic (see Davies’ Arithmetic, § 22), that 
the value of a product is not affected by changing the order of its 
factors: that is, 

12a = ab X12 = ba X12 = ax 12x Bb. 


For convenience, however, the letters in each term are generally 
arranged in alphabetical order, from the left to the right. 

Let it be required to multiply 7a*l? by 4a?d. 

By decomposing the multiplicand and multiplier into their fac- 
tors, we may write the product under the form 


7a3b2 ~ 4a2b = Taaabb x 4aab; 
and since we may change the order of the factors without affect- 
ing the value of the product, we have, 
7a3b? x 4a2b = 7 X 4aaaaabbb = 28258? ; 

a result: which is obtained by multiplying the co-efficients to- 
gether for a new co-efficient, and adding the exponents of the 
same letter, for the new exponents. 

Again: multiply the monomial 12a?b‘c? by 8ab?d?. 

We can place the product under the form, 

12a?btc? x 8a3b?d? = 12 x Baaaaabbbbbbccdd = 96a5b*c?d?. 

By considering the manner in which these results are obtained, 

we see that any quantity, as a, must be found as many times 


28 ELEMENTS OF ALGEBRA. (CHAP. I. 


a factor in the product, as it is a factor in both the multiplicand 
and multiplier; which number will always be expressed by the 
sum of its exponents. 

41. Hence, for the multiplication of monomials we have the 
following 

RULE. 

I. Multiply the co-efficients together for a new co-efficient. 

II. Write after this co-efficient all the letters which enter into the 
multiplicand and multiplier, affecting each with an exponent ua! 
to the sum of tts exponents in both factors. 














EXAMPLES. 
(1) - - 8a2bc2 x Tabd? = 56a3b2c2d2, 
(2) - - 21a5b?de x 8abe? = 168a*b3 etd. 
(3) (4) (5) (6) 
Multiply - —- 3a"b - - 12a’ - - Gayz = = aay 
by -= - 2ba? - - l2e*y - - GPs eee 
Bath? 144a?ay 6ary3z? 2a*x2y?, 
7. Multiply 8a5d?c by Ta®h*cd. Ans. 56a13b7c?d. 
8. Multiply 5abd? by 12cd°. Ans. 60abed'. 
9. Multiply Tathd?c? by abde. Ans. 7a°b?d3ct. 


42. We will now proceed to the multiplication of polynomials. 
In order to explain the most general case, we will suppose the 
multiplicand and multiplier each to contain additive and subtrac- 
tive terms. 

Let a represent the sum of all the additive terms of the multi- 
plicand, and b the sum of the subtractive terms; ¢ the sum of 
the additive terms of the multiplier, and d the sum of the sub- 
tractive terms. The multiplicand will then be represented by 
a —b and the multiplier, by ¢ — d. 

We will now show how the multiplication expressed by 
(a — b) x (ce —d) can be effected. 

The required, product is equal to a —b a —b 





taken as many times as there are units ec —d 

in c—d. Let us first multiply by ¢; ac — be 

that is, take a@—b as many times as — ad + bd 
there are units inc. We begin by wri- ac ai beldsied ibd. 


ting ac, which is too great by 4 taken 


- CHAP. II.] MULTIPLICATION. 29 


¢ times; for it is only the difference between a and 4, that is first 
to be multiplied by c. Hence, ac — be is the product of a —b 


by c. But the true product is a —b taken c —d times: hence, 
the last product is too great by a —6 taken d times; that is, 
by ad — bd, which must be subtracted. Changing the signs and 
subtracting this from the first product (Art. 37), we have 
(a — b) x (e — d) = ac — be — ad + bd: 
If we suppose a and c¢ each equal to 0, the product will re- 


duce to + dd. 


43. By considering the product of a—b by ec —d, we may 
deduce the following rule for the signs, in the multiplication of 
two polynomials. 

When two terms of the multiplicand and muliiplier are affected 
with the same sign, their product will be affected with the sign +, 
and when they are affected with contrary signs, their product will be 
affected with the sign —. 

Again, we say in algebraic language, that + multiplied by +, 


or — multiplied by —, gives + ; — multiplied by +, or + mul- 


tiplied by —, gives —. But since mere signs cannot be multi- 


plied together, this last enunciation does not, in itself, express a 
distinct idea, and should only be considered as an abbreviation 
of the preceding. 

This is not the only case in which algebraists, for the sake of. 
brevity, employ expressions in a technical sense in order to se- 
cure the advantage of fixing the rules in the memory. 


44. Hence, for the multiplication of polynomials we have the 


following 
RULE. 


Multiply ali the terms of the multiplicand by each term of the 


multiplier in succession, affecting the product of uny two terms with 


the sign plus, when their signs are alike, and wit the sign minus, 


; 





when their signs are unlike. Then reduce the polynomial result to 
as simplest form. 
1. Multiply - - - - - - 3a?+ 4ab+ 0? 
by - - = = =. + + 2a + 5d 
6a3 + 8a2b-+ 2ab? 


- The product after reducing, + 15a?b + 20ab? + 563 


becomes - - - - - 6a? + 23a) + 220)? + 5b), 


30 ELEMENTS OF ALGEBRA. [CHAP. Il. 


(2). , (3). 





x? + y? 5 + ay + Tax 
rome y ax + 5ax 
x? 4 + xy? “ax® + ax*y8 +- Ta*x? 
— ay”? — 73 + 5ax® + San?y® + 35a2x? 
e+o — y? 6ax® + bax*y® - 420722, my 


4. Multiply a? + 2ax + a? by aw+a. 
Ans. «3 + 3ax? + 3a2x + a3. 


5. Multiply a?-+ y? by a+y. 
Ans. 3 + xy? + ay + ¥3. 


6. Multiply 3ab? + 6a%c? by 3ab? + 3ac?, 
. Ans. 9a2b* + 27a3b?c? + 18atct. 


7. Multiply 42?— 2y by 2y. Ans. 8x*y — 4y?. 
8. Multiply 2a + 4y by 2x — 4y. Ans. 4x — 16y?. 





9. Multiply «+ ay + ay?+ y3 by «e—y. Ans. 
10. Multiply a? + ay + y? by x2? —ay+ y?*. 
Ans. a* + xy? + y'. 
In order to bring together the similar terms, in the product of 
two polynomials, we arrange the terms of each polynomial with 
reference to a particular letter. 


11. Multiply 4a3— 5a7b — 8ab? + 203 
Dey, 2a2 — 3ab — 40? 
8a5 — 10a4b — 1603S? + 40% 
— 12a*h + 15a°b? + 24a7b3 — 6abt 
— 16a°b? + 20a7b? + 32ab* — 8b 
8a° — 22atb — 17a%b? + 48a7b3 + 26abt — 85. 
After having arranged the polynomials, with reference to the 
letter a, multiply each term of the first, by the term 2a? of the 
second; this gives the polynomial 8a°— 10ath — 16a5b? + 4753, 
the signs of which are the same as those of the multiplicand. 
Passing then to the term — 3ab of the multiplier, multiply each 
term of the multiplicand by it, and as it is affected with the 
sion —, affect each product with a sign contrary to that of the 
corresponding term in the multiplicand; this gives 


— 12a*b +- 15a°b? + 24a2b3 — 6ab4 


- 





CHAP. II.) MULTIPLICATION. 31 


The same operation is also performed with the term — 40?, 


which is also subtractive; this gives, 


— 16a3b? + 20a7b3 + 32ab+ — 865. 
The product is then reduced, and we finally obtain, for the. most 
sumple expression of the product, 
8a>5 — 22ath — 17a%b? + 48a2b3 + 26ab4 — 8b5. 
12. Multiply 2a? — 3ax + 4a? by 5a? — bax — 227... 
Ans. 10a* — 27a3x + 34a?2x? — 18ax3 — Sat. 
13. Multiply 32? —2yx+5 by a2? -+ 2ay —3. 
Ans. 3x* + 4a3y — 4x? — 4a?y? + l6xy — 15. 
14. Multiply 3a? + 2x°y? + 3y? by 2a? — 3a°y? + 5y?. 
Bee. ; 6x® — Sa®y? — 6aty* + Gary? + 15ax3y3 
— Qx?y* + 10x*y5 + 15y5. 
15. Multiply 8axr —6ab—c by 2axr+ab+e. 
Ans. 16a?x? — 4a2bx — 6a2b? + 6acx — Tabe — c?. 
16. Multiply 3a? — 5b? -+ 3c? by a* — 6. 
Ans. 3a* — 5a?*b? + 3a2c? — 3a2b3 + 5b5 — 3)3¢?- 
17. Multiply 3a? — 5bd+ cf 
by — 5a® + 4bd — 8cf 


Prod. red. — 15at + 37a%bd — 29acf — 200?d? + 44bedf — 8c?f?. 











18. Multiply 40%)? — 5a2b?2c + 8a*bc? — 3a?c3 — Tabc3 
by Zab? —4abe —2dc? + 03. 
8atd+ — 10a%bte + 28a3b%c2 — 34a3b2c3 
Prod. red. 





— 4a?b3c* — 16atb3c + 12a%bct + 7a2b2ct 
+ i4a*be? + l4ab*c® — 3a%c? ~ — Tabc®. 





45. Results deduced from the multiplication of polynomials. 

Ist. If the polynomials which are multiplied together are ho- 
mogeneous, 

Their product will also be homogeneous, and the degree of each 
term will be equal to the sum of the degrees of any two terms 
of the multiplicand and multiplier. 

Thus, in example 18th, each term of the multiplicand is of 
the 5th degree, and each term of the multiplier of the 3d de- 
gree: hence, each term of the product is of the 8th degree. 
This remark serves to discover any errors in the addition of the 
exponents. 


82 ELEMENTS OF ALGEBRA. [CHAP. Il 


2d. If no two of the partial products are similar, there will 
be no reduction among the terms of the entire product: hence, 

The total number of terms in the entire product will be equal to 
the number of terms in the multiplicand multiplied by the number 
of terms in the multiplier. 

This is evident, since each term of the multiplier will produce 
assmany terms as there are terms in the multiplicand. ‘Thus, in 
example 16th, there are three terms in the multiplicand and two 
in the multipher: hence, the number of terms in the product is 
equal to 3.x 2= 6. 

3d. Among the different terms .of the product, there are always 
some which cannot be reduced with any others. For, let us 
consider the product with reference to any letter common to the 
multiplicand and multiplier. ‘Then, the irreducible terms are, 

Ist. The term produced by the multiplication of the two terms 
of the multiplicand and multiplier which contain the highest ex- 
ponent of this letter; and the term produced by the multiplica- 
tion of the two terms which contain the lowest exponent of this 
letter. For, these two partial products will contain this letter, 
affected with a higher and lower exponent than either of the 
other partial products. and consequently, they cannot be similar to 
any of them. ‘This remark, the truth of which is deduced from 
the law of the exponents, will-be very useful in division. 

Multiply - - 5atb? + 3a2b — abt — 2ab 

BYE = ysl a’b-— ab? 

5a%b3 + 3a'b? — a3b5 — 2adb4 
___— 5a°b*t — 30°63 + a7b§ 4+ 206°. 

If we examine the multiplicand and multiplier, with reference 
to a, we see that the product of 5atb? by a*b, must be irre- 
ducible ; also, the product of — 2ab3 by ab?. If we consider the 
letter b, we see that the product of — ab+ by — ab?, must be 
irreducible, also that of 3a*b by a*b. 


Product, 


46. We will apply the rules for the multiplication of algebraie 
quantities in the demonstration of the following theorems. 


THEOREM I. 


The square of the sum of two quantities 1s equal to the square 
of the first, plus twice the product of the i. by the second, plus 
the square of the second. . 








CHAP. II.] MULTIPLICATION. 33 


Let a denote one of the quantities and 6 the other: then 
a+ 6= their sum. 
Now, we have from known principles, 
(a+ 6)? = (a+ db) X (a+ 5) = a@ + 2ab + B?, 
which result is the enunciation of the theorem in the language 
of Algebra. , 
To apply this result to finding the square of the binomial 


5a? + 8a, 
we have (5a? + 8a?b)? = 25a* + 80atd + 64a4d?. 
Also, (6atb + 9ab?) = 36a°b? + 108a5h* + 814768 ; 
also, (8a? +- Tach)? =. 


THEOREM It. 


The square of the difference between two quantities ts equal to 
the square of the first, minus twice the product of the first by the 
_ Second, plus the square of the second. 

Let a represent one of the quantities and b the other: then 

a—h=— their difference. 

Now, we have from known principles, 

(a — b)? = (a — b) X (a — b) = a? — 2ab + 2B, 
which is the algebraic enunciation of the theorem. 

To apply this to an example, we have 

(7a?b? — 12ab*) = 49ath* — 168a9b5 + 144078. 
Also, (4a°b3 — 7ec*d3)? = 


THEOREM IItf. 


The product of the sum of two quantities multiplied by thetr 
difference, is equal to the difference of their squares. 

Let the quantities be denoted by a and 3d. 

Then, a+6= their sum, and a —) = their difference. 

We have, from known principles, 


(a+b) x (a—b) = 2 — P, 
which is the algebraic enunciation of the theorem. 
To apply this principle to an example, we have 
(8a3 + Tab”) x (8a3 — Tab?) = 64a® — 49074. 
Also, (9a°e + 7ab*) x (9a%e — T7ab*) = | 


oO 


34 ELEMENTS OF ALGEBRA. [CHAP. II 


47. By considering the last three results, it will be perceived 
that their composition, or the manner in which they are formed 
from the multiplicand and multiplier, is entirely dependent of 
any particular values that may be attributed to the letters a@ and b 
which enter the two factors. 

The manner in which an algebraic product is formed from its 
two factors, is called the law of this product; and this law re- 
mains always the same, whatever values may be attributed to the 
letters which enter into the two factors. 


Of factoring Polynomials. 

48. A given polynomial may often be resolved into two factors 
by mere inspection. ‘This is generally done by selecting all the 
factors common to every term of the polynomial for one factor, 
and writing what remains of each term within a parenthesis fos 
the other factor. 

1. Take, for example, the polynomial 

ab + ac; 
in which, it is plain, that @ is a factor of both terms: hence 
ab + ac = a(b+ ec). 

2. Take, for a second example, the polynomial 

abc + 5ab? + ab2c?. 

It is plain that @ and 0? are factors of all the terms : hence 

ab?c + 5ab3 + abc? = ab* (e + 5b + c?), if’ 

3. Take the polynomial 25a* — 30ab + 15a?b? s Nit, is evidgnt 
that 5 and a? are factors of each of the terms. We “may, there- 


: 
fore, put the polynomial under the form 


5a? (5a? — 6ab + 32). e 
4. Find the factors of 3a?b + 9a’e + 18a*xy. | 
Ans. 3a" (b + 3c + 62y). . 
5. Find the factors of 8a’cx — 18acu? + 2achy — 30a%c%x. 
Ans. 2ac (4ax — 9x? + cty — 15a5c8z). 
6. Find the factors of 24a7b?ca — 30a®b'cby + 36a7b8ced + Gabe. 
Ans. 6abe (4aba — 5a™htc5y + 6aeb7d + 1). 
7. Find the factors of a? -+ 2ab + 6?. 
Ans. (a + b) X (a + 3). 


CHAP. II.] DIVISION. 35 


8. Find the factors of a — 3. Ans. (a+ b) x (a — B). 


9. Find the factors of a? — 2ab + 8. 
Ans. (a — 6) xX (a — Dd). 
10. Find the factors of the polynomial 6a%b + 8a?b5 — 16a’ 
— 2ab. 


11. Find the factors of the polynomial 15abc? — 3bc? + 9a3b5c® 
— 12db%c?, 
12. Find the factors of the polynomial 25a%dc® — 30a%bc*d 
- 5act — 60ac®. 
13. Find the factors of the polynomial 42a?b? — Tabed + 7abd. 
Ans. Tab (6ab — cd + d). 


———— 


DIVISION. 


49. Diviston, in Algebra, explains the method of finding from 
two given quantities, a third quantity, which multiplied by the 
first shall produce the second. 

The first of the given quantities is called the divisor: the sec- 
ond, the dividend; and the third, or quantity sought, the quotzent. 

Let us first consider the case of two monomials, and divide 
35a°b2c by Tab. 

The division may be indicated thus, 

lala eaihaumben 't? =aoatvee. 
Tab . 

Now, since the quotient must be such a quantity as multiplied 
by the divisor will produce the dividend, the co-efficient of the 
yuotient multiplied by 7 must give 35, the co-efficient of the 
lividend; hence, the new co-efficient 5 is found by dividing 35 
vy 7. Again, the exponent of any letter, as a, in the quotient, 
udded to the exponent of the same letter in the divisor, must 
xive the exponent of this letter in the dividend: hence, the ex- 
yonent in the quotient is found by subtracting the exponent in 
he divisor from that in the dividend. Thus, the exponent of 


gis ee 2 and of by¢9%)— T.—='1,. 
ind since c is not found in the divisor, there is nothing to be 
subtracted from its exponent. 





50. Hence, for the division of monomials, we have the following 


36 ELEMENTS OF ALGEBRA. [CHAP. II. 


RULE. 
I. Divide the co-efficient of the dividend by the co-efficient of the 
divisor, for a new co-efficient. 
Il. Write after this co-efficient, all the letters of the dividend, 
and affect each with an exponent equal to the excess of its exponent 
in the dividend over that in the divisor. 


From this rule we find, 


facie = 4a*bed$ a = 5a*beed. 
1, Divide 16x? by 8x. Ans. 2x 
2. Divide 15a’xy? by 3ay. — Ans. Saxy?. 
3. Divide 84ab3x by 1207. Ans. Tabx. 
4. Divide 96atb?c? by 12a7be. Ans. 8a*he?. 
5. Divide 144a%b8c7d° by 36ath®c®d. Ans. 4a5b?eds, 
6. Divide 256a%bc?x? by 16a?cx?. Ans. 16abce. 
7. Divide 300a5b*c3x? by 30atb3c*x. Ans. 10abex. 


51. It follows from the preceding rule that the exact division 
of monomials will be impossible. 

Ist. When the co-efficient of the dividend is not divisible by 
that of the divisor. 

2d. When the exponent of the same letter is greater in the 
divisor than in the dividend. 

3d. When the divisor contains one or more letters which are 
not found in the dividend. 

When either of these three cases occurs, the quotient remains 
under the form of a monomial fraction; that is, a monomial ex- 
pression, necessarily containing the algebraic sign of division. 
Such expressions may frequently be reduced. 
12a‘*b?ed = 33a®bd 
| vaghee > ate 

Here an entire monomial cannot be obtained for a quotient; 
for, 12 is not divisible by 8, and moreover, the exponent of ¢ 
is less in the dividend than in the divisor. But the expression 
can be reduced, by dividing the numerator and denominator by 
the factors 4, a2, b, and c, which are common to both the terms 
of the’ fraction. 


Take, for example, 


CHAP. II] DIVISION. 37 








In general, to reduce a monomial fraction, we have the fol- 


2 lowing 


RULE. 


Suppress all the factors common to the numerator and denomina- 
tor, and write those letters which are not common, with their respec- 
tive exponents, in the term of the fraction which contains them. 


From this rule we find, 





48a3b2cd3 4ad? 37ab3cid 37b2c 
SS AIEEE IED LA ee 
; 16a7b¢9 Ac ’ 14a3b2 ~ 2ab’ 


In the last example, as all the factors of the dividend are 
found in the divisor, the numerator is reduced to unity; for, m 
fact, both terms of the fraction are divided by the numerator. 


52. It often happens, that the exponents of certain letters, are 


the same in the dividend and divisor. 


24a3h2 
gaegay en 


is a case in which the letter b is affected with the same expo- 


For example, - 


nent in the dividend and divisor: hence, it will divide out, and 


will not appear in the quotient. 
But if it is desirable to preserve the trace of this letter in 
the quotient, we may apply to it the rule for the exponents (Art. 


50): which gives 


j2 
ori p22 Fo 
rg 62-2 — b 
This new symbol 5°, indicates that the letter 6 enters 0 times 
as a factor in the quotient (Art. 13); or what is the same thing, 
that it does not enter it at all. Still, the notation shows that b 


was in the dividend and divisor with the same exponent, and has 


disappeared by division. 


15a?b3c? 


Dkather, pated 5ab2c9 soa 562. 


In like manner, 


53. We will now show that the power of any quantity whose 


_ exponent is 0, is equal to unity. Let the quantity be represented 
| by a, and let m denote any exponent whatever 


38 ELEMENTS OF ALGEBRA. {CHAP. II. 


Then, — =a" = a, by the rule,of division, 
a 
a™ : ‘ 
But, — = 1, since the numerator and denominator are equal: 
a 
: 4 a™ 
hence, a® = 1, since each is equal to —. 
a 


We observe again, that the symbol a@® is only employed con- 
ventionally, to preserve in the calculation the trace of a letter 
which entered in the enunciation of a question, but which may 
disappear by division. 


Division of Polynomials. 


54. The object of division, is to find a third polynomial called 
the quotient, which, multiplied by the divisor, shall produce the 
dividend. 

Hence, the dividend is the assemblage, after reduction, of the 
partial products of each term of the divisor by each term of the 
quotient, and consequently, the signs of the terms in the quotient 
must be such as to give proper signs to the partial products. 

Since, in multiplication, the product of two terms having the 
same sign is affected with the sign +, and the product of two 
terms having contrary signs, with the sign —, we may conclude, 

Ist. That when the term of the dividend has the sign +, and 
that of the divisor the sign of +, the term of the quotient must 
have the sign -}. 

2d. When the term of the dividend has the sign +, and that 


of the divisor the sign —, the term of the quotient must have 
the sign — ; because it is only the sign —, which, combined with 
the sign —, can produce the sign + of the dividend. 


3d. When the term of the dividend has the sign —, and that 
of the divisor the sign +, the quotient must have the sign —. 


That is, when the two corresponding terms of the dividend 
and divisor have the same sign, their quotient will be affected 
with the sign +, and when they are affected with contrary signs, 
their quotient will be affected with the sign —; again, for the 
sake of brevity, we say that 

+ divided by +, and — divided by —, give +; 
— divided by +, and + divided by —, give —. 


CHAP. II1.] DIVISION. 39 


FIRST EXAMPLE. 
Divide a? —2ax-+ «a? by a—-z. 
It is found most convenient, Dividend. Divisor. 


in division in algebra, to place a — 2an + # || a—@ 
the divisor on the right of the a — a ‘eee 
dividend and the quotient di- — get a?  Nyotent. 
rectly under the divisor. — an + x. 


We first divide the term a? of the dividend by the term a of 
the divisor, the partial quotient is a, which we place under the 
divisor. We then multiply the divisor by a, and subtract the 
product a? — ax from the dividend, and to the remainder bring 
down «x. We then divide the first term of the remainder, — aa, 
by a, the quotient is — a. We then multiply the divisor by 
— «x, and, subtracting as before, we find nothing remains. Hence, 
a—wx is the exact quotient. 


SECOND EXAMPLE. 


Let it be required to divide 26a7b? + 10at — 480°) + 24ab? by 
4ab — 5a + 302. In order that we may follow the steps of the 
operation more easily, we will arrange the quantities with refer- 
ence to the letter a. 

Dividend. Divisor. 
10a* — 48a3b + 26a2b? + 24ad3 — 5a? + 4ab + 306? 
+ 10at— 809 — 6a2b? — 2a? + 8ab 
— 4003) + 32020? + 24ab> Quanent 
— 40a°b + 3202b? + 24063. 











It follows from the definition of division, and the rule for the 
multiplication of polynomials (Art. 44), that the dividend is the 
assemblage, after addition and reduction, of the partial products 
of each term of the divisor, by each term of the quotient sought. 
Hence, if we could discover a term in the dividend which was 
derived, without reduction, from the multiplication of a term of 
the divisor by a term of the quotient, then, by dividing this term 
of the dividend by that of the divisor, we should obtain a term of 
the required quotient. 

Now, from the third remark of Art. 45, the term 10a‘, affected 
with the highest exponent of the letter a, is derived, without re- 
duction from the two terms of the divisor and quotient, affected 


— 


40 ELEMENTS OF ALGEBRA. [CHAP. If. 


with the highest exponent of the same letter. Hence, by dividing 
the term 10a* by the term — 5a”, we shall have a term of the 
required quotient. 


Dividend. Divisor. 


10a* — 48a3b + 26a2b2 +. 24ab3 | — 5a? +. 4ab + 35? 





+ 10at*— 8a°b — 6a?b? — 2a’ a 8ab 
— 40a3b + 320%)? + 24ab3 wena 
— 40a3b + 32a?b? + 24a63. 
Since the terms 10a* and — 5a? are affected with contrary 
signs, their quotient will have the sign —; hence, 10a‘, divi- 


ded by — 5a’, gives — 2a? for a term of the required quotient. 
After having written this term under the divisor, multiply each 
term of the divisor by it, and subtract the product, 


10a* — 8a*b + 6a?d?, 


from the dividend, which is done by writing it below the divi- 
dend, conceiving the signs to be changed, and performing the re- 
duction. ‘Thus, the remainder after the first partial division is 


— 40a3b + 32a2b?2 + 24ab3. 


This result is composed of the partial products of each term 
of the divisor, by all the terms of the quotient which remain to 
be determined. We may then consider it as a new dividend, and 
reason upon it as upon the proposed dividend. We will there- 
fore divide the term — 40a%), affected with the highest exponent 
of a, by the term — 5a? of the divisor. Now, from the prece- 
ding principles, 

— 40a%b, divided by — 5a’, gives + 8ab 
for a new term of the quotient, which is written on the right of 
the first. Multiplying each term of the divisor by this term of 
the quotient, and writing the products underneath the second divi- 
dend, and making the subtraction, we find that nothing remains. 
Hence. 
— 2a2-+ 8ab or 8ab — 2a? 


is the required quotient, and if the divisor be multiplied by it, 
the product will be the given dividend. 

By considering the preceding reasoning, we see that, in each 
partial operation, we divide that term of the dividend which is 





GHAP. II.) DIVISION. 41 


affected with the highest exponent of one of the letters, by that 
term of the divisor affected with the highest exponent of the 
same letter. Now, we avoid the trouble of looking out these 
terms by writing, in the first place, the terms of the dividend and 
divisor in such a manner that the exponents of the same letter shali 
go on diminishing from left to right. 

This is what is called arranging the dividend and divisor with 
reference to a certain letter. By this preparation, the first term 
on the left of the dividend, and the first on the left of the divisor, 
are always the two which must be divided by each other in or- 
der to obtain a term of the quotient. 


55. Hence, for the division of polynomials we have the fol- 
lowing . 

; RULE. 

I. Arrange the dividend and divisor with reference to a certain 
letter, and then divide the first term on the left of the dividend by 
the first term on the left of the divisor, for the first term of the 
quotient ; muliply the divisor by this term and subtract the prod- 
uct from the dividend. 

Il. Then divide the first term of the remainder by the first term 
of the divisor, for the second term of the quotient; multiply the 
divisor by this second term, and subtract the product from the re- 
sult of the first operation. Continue the same process, and if the 
remainder is 0, the division is said to be exact. 


THIRD EXAMPLE. 

Divide 2lx%y? + 25x%y3 + 68ayt — 40y° — 56x° — 182xty by 
Sy? — 8x? — Gry. } 
— 40y° + 68xy4 + 25x7y3 + 2lary? — 18aty — 56x° \|5¥” — 6xy — 8x? 
— 407° + 48xy* + 64x77? — 8y3 + 4ay? — Ba2y + 7x? 
Ist rem 20xy* — 39a?y3 + 2123y? 

20xy* — 24x23 — 32x3y? 





2d rem. - —15x?y3 + 53a3y? — 18xty 
— 15x2y3 + 18x3y? + 24aty 
S@ rem. - - - = 35a3y? — 42aty — 5625 


35x%y? — 42aty — 5695 


Final remainder ot TS eal atl Pes > OL 





42 ELEMENTS OF ALGEBRA. [CHAP. Il. 


56. Remarx.—In performing the division, it 1s not necessary 
to bring down all the terms of the dividend to form the first re- 
mainder, but they may be brought down in succession, as in the 
example. 

As it is important that beginners should render themselves 
familiar with the algebraic operation, and acquire the habit of 
calculating promptly, we will treat this last example in a different 
manner, at the same time indicating the simplifications which 
should be introduced. These, consist in subtracting each partial 
product from the dividend as soon as this product is formed. 


— 4075 + 68ay* + 25x7y3 + 21 x3y? — 18aty — 56x° | 5y? — 6ry — 8x? 





Ist rem. 20xy* — 39x?y3 + 21a%y? — 8y3 + dary? — 3a2y + 72x 
2d rem. - —15a?y3 + 53a3y? — 18aty 

3d rem. - - = — 385x3y? — 42aty — 56x° 

Final remainder - - - - - 0. 


First, by dividing — 40y3 by 5y?, we obtain — 8y° for the quo- 
tient. Multiplying 5y? by — 8y?, we have — 40y°, or by chan- 
ging the sign, + 40y5, which destroys the first term of the divi- 
dend. 

In like manner, — 6ry x — 8y° gives + 48zy*, and for the 
subtraction — 48axy*, which reduced with + 68ay*, gives 20ry* 
for a remainder. Again, — 8x? x — 8y? gives +, and changing 
the sign, — 64x?y3, which reduced with 2527y3, gives — 397? 
Hence, the result of the first operation is 20ay* — 39x?y3, fol- 
lowed by those terms of the dividend which have not been re- 
duced with the partial products already obtained. For the sec- 
ond part of the operation, it is only necessary to bring down the 
next term of the dividend, to separate this new dividend from the 
primitive by a line, and to operate upon this new dividend in 
the same manner as we operated upon the primitive, and so on. 


FOURTH EXAMPLE. 
Divide - - 95a — 73a? + 56a* — 25 — 59a? by — 3a? 
+5 — lla — 7a. 
560**— 594° ~— 73? + 95a — 25 | 7a — 3a — lla +5 
{st rem. — 35a3+ 15a? + 55a — 25 8a —95 


Qd remainder - - 0. 








CHAP. II.] DIVISION. 43 


GENERAL EXAMPLES. 
1. Divide 10ab + 15ac by 5a. Ans. 2b + 3e. 
2. Divide (30axz — 54x by 6a. Ans. 5a — 9. 
3. Divide 10x?y — 15y?— 5y by 5y. Ans. 2a? — 3y—1. 
4. Divide 12a@ + 3a% — 18ax? by 3a. Ans. 4 + « — 62”. 
5. Divide 6ax?+ 9a?x + aa? by az. Ans. 62 + 9a + az. 
\6. Divide a? + 2ax + a? by a+ cz. Ans. a+. 
7. Divide a? — 3a’y + 3ay? — y? by a— y. 
Ans. a® — 2ay + ¥? 
8. Divide 24a7b — 12a%ch? — 6ab by — 6ab. 
Ans. — 4a + 2a2ch + 1. 
9. Divide 621 —96 by 3r—6. Ans. 2x°+ 4u?+ 82+ 16. 
7 10. Divide - - = @ — 5ata + 10a3x? — 10a2x? + 5ax* — a 
by a — 2ax 4 2. Ans. a3 — 3a%x + 3ax? — x. 
1f. Divide 48x3 — 76ax? — 64a2x + 105a° by 2x — 3a. 
Ans. 24x”? — 2ax — 35a?. 
12. Divide y® — 3y*a? + 3y?xt — a6 by y3 — 3y2a + 3yx? — 2%. 
Ans. y? + 3y?x + Sya? + x3, 
13. Divide 64a*b® — 25a2b8 by 8a7b? 4 5abt. 
Ans. 8a7b? — 5ab+. 
14. Divide 6a3 + 23a7b + 22ab? + 503 by 3a? + 4ab + 0. 
Ans. 2a + 56. 
15. Divide 6aa® + aay 6+ 42a%a? by ax + Sax. 
Ans. 0° -+- xy® + tae 
16. Divide — 15a*+ 37a?bd — 29a*cf — 20b?d? + 44bcedf — 8c?f? 
by 3a? —,5bd + ef. Ans. — 5a? + 4bd — 8ef. 
17. Divide at + ay? + y4 by a? — ay + y?. 
| Ans, «® + ry + y?. 
»18. Divide at —y* by a—y. Ans. «3 + a?y + ay? + y?. 
19. Divide 3at — 8a?b? + 3a2c? + 5b4 — 3b?c? by a? — b?. 
Ans. 3a? — 5b? + 3c?. 
20. Divide 6x°— 5x5y?— 6atyt + 6ax3y? + 15a3y? — 9x?y* + 10x?y% 
+ 15y® by 3x? + 2x2y? + 3y?. Ans. 203 — 3x?y? + 5y% 


44 ELEMENTS OF ALGEBRA. (CHAP. II. 


Remarks on the Division of Polynomaals. 


57. When the first term of the arranged dividend is not ex- 
actly divisivle by that of the arranged divisor, the complete divis- 
ion is impossible; that is to say, there is not a polynomial 
which, multiplied by the divisor, will produce the dividend. And 
in general, we shall find that a division is impossible, when the 
first term of any one of the partial dividends is not divisible by 
the first term of the divisor. 

We will add, as to polynomials, that it may often be discov- 
ered by mere inspection that they are not divisible. When the 
polynomials contain two or more letters, observe the two terms 
of the dividend and divisor, which are affected with the highest 
exponent of each of the letters. If these terms do not give an 
exact quotient, we may conclude that the total division is im- 
possible. 

Take, for example, 


12a? — 5a2b + Tab? — 1163 || 4a? + 8ab w 362 


By considering only the letter a, the division would appear 
possible ; but regarding the letter b, the division is impossible, 
since — 116° is not divisible by 30. 

58. One polynomial A, cannot be divided by another B con- 
taining a letter which is not found in the dividend; for, it is 
unpossible that a third quantity, multiplied by B which contains _ 
a certain letter, should give a product independent of that letter. 

A monomial is never divisible by a polynomial, because every 
polynomial multiplied by either a monomial or a polynomial gives: 
a product containing at least two terms which are not suscep- 
tible of reduction. 


59. If the letter, with reference to which the dividend is ar- 
ranged, is not found in the divisor, the divisor is said to be in- 
dependent of. that letter; and in that case, the exact division is 
impossible, unless the divisor will divide separately the co-efficient 
of each term of the dividend. 

For example, if the dividend were 


Bhat + 9ba? + 128, 


arranged with reference to the letter a, and the divisor 30, the 


CHAP. ITI.) DIVISION. 45 


divisor would be independent of the letter a; and it is evident that 
the exact division could not be performed unless the co-efficient 
of each term of the dividend were divisible by 3b. The expo- 
nents of the leading letter in the quotient would be the same as 
in the dividend. 


1. Divide 18a°x? — 36a?x3 — 12axr by 6x. 
; Ans. 3a3x — 6a?x? — 2a. 


2. Divide 25atb — 30a*b + 40ab by 5b. 
Ans. 5a* — 6a? + 8a. 


60. :Although there is some analogy between arithmetical and 
algebraical division, with respect to the manner in which the 
operations are disposed and performed, yet there is this essential 
difference between them, that in arithmetical division the figures 
of the quotient are obtained by trial, while in algebraical division 
the quotient obtained by dividing the first term of the partial divi- 
dend by the first term of the divisor, is always one of the terms 
of the quotient sought. ‘ 

From the third remark of Art. 45, it appears that the term of 
the dividend affected with the highest exponent of the leading 
letter, and the term affected with the lowest exponent of the 
same letter, may each be derived without reduction, from the 
multiplication of a term of the divisor by a term of the quotient. 
Therefore, nothing prevents our commencing the operation at the 
right instead of the left, since it might be performed upon the 
terms affected with the lowest exponent of the letter, with ref- 
erence to which the arrangement has been made. 

Lastly, so independent are the partial operations required by 
the process, that after having subtracted the product of the di- 

isor by the first term found in the quotient, we could obtain 
another term of the quotient by dividing by each other the two 
terms of the new dividend and divisor, affected with the highest 
exponent of a different letter from the one first selected. If the 
"game letter is preserved, it is only because there is no reason 
for changing it, and because the two polynomials are already 
arranged with reference to it; the first terms on the left of the 
dividend and divisor being sufficient to obtain a term of the 
quotient ; whereas, if the letter is changed, it would be neces- 
sary to seek again for the highest exponent of this letter 


46 ELEMENTS OF ALGEBRA. [CHAP. IJ 


61. Among the different examples of algebraic division, there 
is one remarkable for its applications. It is expressed thus: 


The difference between the same powers of any two quantities is 
always divisible by the difference between the quantities. 


Let the quantities be represented by a and 5; and let m de- 
note any positive whole number. ‘Then, 


am — hm 
will express the difference between the same powers of a and 3, 
and it is to be proved that a — b™ is exactly divisible by a — b. 
If we begin the division of a™— 0b™ by a—b, we have 
a™ — hm a—b 
Ist rem. Se de ite ee hie bon) Se 
or, by factoring - - - b(a™—1 — hm-}), 


Dividing a” by a the quotient is a"~}, by the rule for the ex- 
ponents. ‘The product of a—b by a™=' being subtracted from 
the dividend, the first remainder is a™—!1) — 6", which can be 
put under the form 8 (a7! — hm), 

Now, if the factor (a"~1 — }™-') of the remainder, be divisi- 
ble by a — 8, it follows that the dividend a™ — J” is also divisi- 
ble by a— 2: that is, . 

If the difference of the same powers of two quantities be divisi- 
hle by the difference of the quantities, then, the difference of the 

powers of a degree greater by unity 2s also divisible by it. 
But by the rules for division, we have 
A See 9B 
a—b 

Hence, we know, from what has just been proved, that a? — 63 
is divisible by a—J, and from that result we conclude that 
at — }* is divisible by a—%, and so on, until we reach any 
exponent at m. 


=a-+t b. 


‘CHAP. IIS.) ALGEBRAIC FRACTIONS. 47 


CHAPTER lI. 
OF ALGEBRAIC FRACTIONS. 


62. AtceEsBratc fractions are to be considered in the same 
‘point of view as arithmetical fractions; that is, @ unit is sup- 
posed to be divided into as many equal parts as there are units in 
the denominator, and one of. these parts is supposed to be taken 
as many times as there are units in the numerator. 

Thus, in the fractional expression 


a+b 

c+d 

a given unit is supposed to be divided into as many equal parts 
as there are units in c+ d, and as many of these parts are 
taken, as there are units in a+ bd. 

The rules for performing Addition, Sato Multiplication, 
and Division, are the same as in arithmetical fractions. Hence, 
it will not be necessary to demonstrate these rules, and in their 
application we must follow the methods already indicated in sim- 
ilar operations on entire algebraic quantities. 


Ld 


_ 63. Every quantity which is not expressed under a fractional 
form, is called an entire algebraic quantity. 


64. An algebraic expression, composed. partly of an_ entire 
quantity and partly of a fraction, is called a mixed quantity. 


65. When the division of two monomial quantities cannot be 
performed exactly, it is indicated by means of the known sign, 
and in this case, the quotient is presented under the form of a 
fraction, which we have already learned how to simplify (Art. 51). 

With respect to polynomial fractions, the following are cases 
which are easily reduced. 


‘ 


48 ELEMENTS OF ALGEBRA. {CHAP. 131. 











a? — 
Take, a example, the expression oe 
. a+ 6)(a— 
This fraction can take the form ( Art. 46). 
ea Hiead « 
Suppressing the factor a — >. which is common to the two terms, 
b 
we obtain 4. [ 
a—b 


: : 3 — 10a?b + 5ab? 
Again, take the expression se PA Sei 


which can be put under the form (Art. 48): 


5a (a? — 2ab + 0b?) 
8a? (a — b) i 











a 5a (a — b)? 

which is equal to Seca : 
and by suppressing the common factors, a(a — bd), the result is 

5(a—'b). 

8a ; 


In the particular cases examined above, the two terms of the 
fraction are decomposed into factors, and then the factors com- 
mon to the numerator and denominator are cancelled. Practice 
teaches the manner of performing these decompositions, when 
they are possible. 

But the two terms of the fraction may be Sern oatea poly- 
nomials, and then, their decomposition into factors not being so 
easy, we have recourse to the process for finding the greatest 
common divisor, which is explained at page 300. 


CASE I. 


70. To reduce a fraction to its simplest form. 


RULE. 
I. Decompose the numerator and denominator into factors, as im 
Art. 48. 
Il. Then cancel the factors common to the numerator and de- 
nommator, and the result will be the simplest form of the fraction. 


. CHAP. III.) ALGEBRAIC FRACTIONS. 49 


EXAMPLES. 


3ab + 6ac 


1. Reduce the fraction ade aa 


to its simplest form. 


We see, by inspection, that 3 and a are factors of the nuv- 


merator, hence 


3ab + 6ac = 3a(b+ 2c) 
We-also see, that 3 and a are factors of the denominator, hence 
Bad + 12a = 3a(d + 4) 
3ab + bac _ 3a (b + 2c) b + 2c 








Meee iis sat 47 aha 
6a2b +- 3ac ie ee 
2. Reduce Qab a Bad to its simplest form. 
, 2ab+e 
ns. cr eraRe 
Q5be+ 5bf 
3. Reduce 350? + 158 to its lowest terms. 
5e + f 
Ans. +3 
54abe : Pt hie 
4. Reduce Iba 4 Oacd to its simplest form. 
65 
Aas. ary 
6a? 12ab : , 
5. Reduce paper. 1 920/ : — ae to its simplest form. ; 
3a + 
Ans. 7h 
12acd — 4cd? , - 
6. Reduce T2edf | ded to its simplest form. 
' 3a—d 
Ans, 
3f+e 


$2 
eee Sof to its simplest form. 


7. hedice = ____ 
27ac2 — 6ac3 f 
6ac — 
7S ag as ee 
iti 9c — 2c? 
CASE II. 


71. To reduce a mixed quantity to the form of a fraction. 
4 


50 ELEMENTS OF ALGEBRA. [CHAP. IIT. 


RULE. 


Multiply the entire part by the denominator of the fraction: then 
connect this product with the terms of the numerator by the rules 
for addition, and under the result place the given denominator. 


























EXAMPLES. 
ill am 
1. Reduce «x — (af oth to the form of a fraction. 
x 
a. Wie Uae _ 2 (a ee et 
x x x 
2 
2. Reduce — eee to the form of a fraction. 
ax — x? 
Ans. a 
22 — 7 - 
3. Reduce 5 + to the form of a fraction. 
ns — 
Ans. tial 
3x 
x—a—l s 
4. Reduce 1 — — to the form of a fraction. 
a 
2a — 1 
hee ark: 
a 
5. Reduce 1 + 22 — 2 to the form of a fraction. 
10x? + 4x + 3 
Ans. f 
5x 


to the form of a fraction. 





6. Reduce 32 —1 — 
3a 


Sax — 4a — Tx + 2 


A 
hd 3a —2 


CASE Iii. 


72. To reduce a fraction to an entire or mixed quantity. 


RULE. 


Divide the numerator by the denominator for the entire part, and 
place the remainder, tf any over the denominator for the fractional 


“part. 


CHAP. III.] ALGEBRAIC FRACTIONS. 51 





EXAMPLES. 
ax + a? ' 
1. Reduce Nad to a mixed quantity. 
ax + a? a 
Ans. oe sae =a+—. 
oe x 
ax — x” ; 
2. Reduce ———— to an entire or mixed quantity. 
ge 
Ans. a — 2. 
ab — 2a? ; 
3. Reduce os to a mixed quantity. 
2a? 
Ans. a — —. 
b 
a? — x ‘ ; 
4. Reduce —M—— to an entire quantity. 
a— 2 
Ans. a+ @. 
wae: 
5. Reduce ———~ to an entire quantity. 
e—y 
Ans. «2 + ay + y?. 
10x? — 5x+ 3 : ‘ 
6. Reduce aa a to a mixed quantity. 
ie 


3 
Ans. 2x —1--+ —. 
ox 
CASE IV. 
73 To reduce fractions having different denominators to equiv 
alent fractions having a common denominator. 
} RULE. 


Multiply, each numerator into all the denominators except tts own, 
for the new numerators, and all the denominators together for a 
common denominator. 


EXAMPLES. 


a b 4 P y 
1. Reduce — and — to equivalent fractions having a com , 
c 


b 
non denominator. 
1 Sox C= dc 
the new numerators. 
| ae 
and - §&§xe¢e=be the common denominator. 
a a+b : j 
2. Reduce er and ——— to fractions having a common de- 
C 
; ac ab -+- b? 
10ominator. Ans. — and Diane 
be be 
UNIVERSITY OF 


- ILLINOIS LIBRARY 


RBRANA Puapsene, ae 


bs ELEMENTS OF ALGEBRA. (CHAP. IIL 


3. Reduce = : a and d, to fractions having a common de- 
a 


3c 
g Scan 4abd 6acd 
nominator. Ans. ——, ——, an : 
6ac 6ac 





Dl mae 2 : 
4. Reduce re > and a+ =, to fractions having a com- 














: 9a «8 12a? + 24x 
mon denominator. Ans. —, a and arial iol 
12a Ila 124 * 
1 a2 a2 + x? : ; 
5. Reduce ose and. weet to fractions having a com- 
mon denominator. ; 
4 3a+ 3a 2a? + are po 6a? +- 6x? 
Ans. —_—, ————__—. 
6a + 62’. 6a +- 6a ~ 6a + 62 
—b b : : 
6. Reduce é : : , and —, to fractions having a com- 
a—b ax c 
mon denominator. 
A acx ac? — abe — be? ++ cb? + a®ba — ab?ax 
ns, ——__——_,__ —___- ———_—__—_+——. 
a’cx — abcx a®cx — abcx : a’cx — abcx 


CASE V. 
74. To add fractional quantities together. 
RULE. 


Reduce the fractions, if necessary, to a common denominator : 
then add the numerators together and place their sum over the 
common denominator. 


EXAMPLES. 
: a c e 
1. Find the sum of Fasege and ri: 
Here, -- a Xd X f= adf 
exbxf=—=cbf 7? the new numerators. 
6X 0 x diss end 
And - bxdx i = oe the common denominator. 
adf cbf A adf + cbf + ebd 
Hence, baf 4 baf ata ST bah the sum. 
2 
2. To hie add b+. 
— Ber? 
Ans. @ 2abx — 3ca 


be ; 


CHAP. III.J ALGEBRAIC FRACTIONS. 53 




















x oo x x 
3. Add es and oe together. Ans. @ +55: 
Lei 4 : a 
4. Add = 5 and — together. Ans. a 
a 22 — 3 
5. Add 3 to 32+ ~ : 
10x — 17 
Ans. 4 ———— 
| ns. 4x + 19 
; 5x? 
6. lt 1s required to add 4a, Oa" and together. 
5x3 : 2 
Angetah ott Ieee 
2ax 
: : oe 2 1 : 
7. It is required to add = > and — i together. 
49x + 12 
Ans. 2 ae. 
ns. 2x0 -+- 60 


: ‘4 . 
8. It is required to add 4z, > and 2+ = together. 





4 
Ans. Ae ba 


: ' 8 
9. It is required to add 3a” + = and «'— = together. 


23a 
Ans. — 
ns. 3x2 + 45 


as 8 c 

a—b a+B ata 

vi a3 — ax? + a®b — bax? + ac + acx — abc —bex + a?d — bd 
RS, ee ee eee eee a Oe ae ee ee ee ee eee 


a? — b2a + ata — bx 


a? + a (b + c + d) — a (a? — cu + bc) — b (x? + cx + bd) 


a + atx — ab? — bx 


10. What is the sum of and 








CASE VI. 


75. To subtract one fractional quantity from another. 


RULE. 


I. Reduce the fractions to a common denominator. 

Il. Subtract the numerator of the subtrahend from the numer- 
ator of the minuend, and place the difference over the common de- 
nominator. 


54 ELEMENTS OF ALGEBRA. (CHAP. IIl 


EXAMPLES. 


— a atl 2a — 42 
2b 3c 





1. Find the difference of the fractions 


Here, (x— a) X 3c = 3cax — 3ac 
the numerators. 
(2a — 4x) x 26 = 4ab — 8ba 


And, 20-x oe == Oe the common denominator. 


3cx — 3ac 4ab—8bx  3cx—3ac— 4ab + 8bx 
bbe WRG be 6be ' 


2. Required the difference of = and ba Ans. Nd 


Hence, 


3. Required the difference of 5y and = Ans. 


: ; 3a 2a 132 
. Required the difference of 7 and ris Ans. $3" 
xta c 
; and Ty 
dx + ad — be 
; bd 


xta a 2a +7 
5b gr 
24x + 8a — 10bx — 35d 


405 


fe 





5. Required the difference of 


Ans 


6. Required the difference of 








Ans 


ex—ea@ 


~3 





. Required the difference of 3x + > and « — 


ce + bx — ab 
Sr sata 


a 


Ans. 22 + 


CASE VII. 


76. To multiply fractional quantities together. 


RULE. ' 
j 


If the quantities to be multiplied are mixed, reduce them to a 
fractional form; then multiply the numerators together for a nu- 
merator and the denominators together for a denominator. 


t 
% 


CHAP. III.] ALGEBRAIC FRACTIONS. 55 











EXAMPLES. 
1. Multiply a+ bi by at 
a d 
a: 
First, - - - pe =e El 
i 3 bd : 
2 2 
anne , a? + bx ? fo" ae + we 
a d ad 
3x 3a 9ax 
2. Required the product of oa and >: Ans. oi 
: 2a 3x? 3x8 
3. Required the product of ee and By Ans. or 
4°Find the continued product of a Biss and eee, 
., a c 2b 
Ans. 9ax. 


5. It is required to find the product of 6+ and — 


ab + be 


x 


Ans. 


an 
and foe 


. Requi ce * 
6. Required the product of ¥ ae 








7. Required the product of «+ as, and) -— : 





8. Required the product of a+ mas 


a—wx Ye aa 


CASE VIII. 
77. To divide one fractional quantity by another. 


RULE. 


Reduce the mixed quantities, if there are any, to a fractional 
form: then invert the terms of the divisor and multiply the frac- 
tions together as in the last case. 


Zill ae 























56 ELEMENTS OF ALGEBRA. (CHAP. III 
EXAMPLES. 
4 | 
1. Divide - - - a@—— by ds 
2c g / 
6 2ac—b 
7h Be 2c . 
6 f 2c—b  g  acg—bg 
Hence, oe eer = es Def 
"Fe of 12 91x 
2. Let = be divided by 2 Ans. rh 
eb4a2 bade r 
3. Let -—— be divided by 5z. Ans. £. 
7 me 
a+l 22 Pa 
4. Tet, 6 be divided by > Ans. 
2 
5. Let mE be divided by > Ans a 
5x - 2a 5ba 
6. Let = be divided by 35° Ans. 3a" 
x—b 416 hues SER x —b 
7. Let Sid be divided by aa" Ans. oc 
at — Of ys 2? -+ be 
8. Let Wags aia Fr be divided by Gas : 
Ans. x +— 
— 1 —xr—1 
9. Divide & a by ee ee pad bo. Ea ae 
—.2 1 — x? a 
10. Divide “*— py 744. Ans. ~ (V4 0 
If we have a fraction of the form ee 
Bigtc 
e soonnts 
we may observe that 
—a a —a : 
= also —=-° and —; Sie; that is, 


The sign of the quotient will be changed by changing the sign 
either of the numerator or denominator, but will not be affected by 
changing the signs of both the terms. 


-CHAP. ITI.] ALGEBRAIC FRACTIONS. 57 


78. We will add but two propositions more on the subject of 
‘fractions. 


If the same number be added to each of the terms of a proper 
fraction, the new fraction resulting from this addition will be greater 
than the first; but if it be added to the terms of an improper 
fraction, the resulting fraction will be less than the first. 


._ 


Let the fraction be expressed by = and suppose a< b. 
Let m represent the number to be added to each term: then 


: sat fi 
2 atm 
the new incon Me ‘i 
ioe 
' Tn order = 


b+ m 
pare the two fractions, they must be reduced 
to the same denominator, which gives for 





a __ab+am 


bb” B+bm A. 


the first fraction, 


atm ab+ bmn 
btm b+ bm’ 


é 

é 
Now, the denominators being the same, that Pa will be 
the greatest which has the greater numeratoy®.,, | se to nu- 
merators have a common part ab, and th t’ bt of gihe sec- 
ond is greater than the part am of the firs , Hinge b Sires yfice 

he e 
ab + bm > ab + am; & 


that is, the second fraction is greater than the first. % 





and for the new fraction, 


If the given fraction is improper, that is, if a>, it is plain 
that the numerator of the second fraction will be less than that 
of the first, since 6m would then be less than am. 


If the same number be subtracted from each term of a proper 
fraction, the value of the fraction will be diminished; but if it be 
subtracted from the terms of an improper fraction, the value of the 
fraction will be increased. 

a 

Let the fraction be expressed by > and denote the number 
to be subtracted by m. 
_ Then, 


a—m : : 
— —  — the new fraction 
b—m 


a -« 


58 ELEMENTS OF ALGEBRA. (CHAP. IIT 


By reducing to the same denominator, we have, ) 


a _ ab — am 
b. .b? ba 


a—m ab —bm ° 
b—-m b—bm 

Now, if we suppose a <b, then am < bm; and if am< bm, 
then will 


and 


ab — am > ab — bm: 
that is, the new fraction will be less than the first. 
If a> 6, that is, if the fraction is improper, ‘then 
am >bm, and ab —am< ab a bm, 
that is, the new fraction will be greater than fie first. 


GENERAL EXAMPLES. 


























1 + 2 Loe 2 (1 + at) 
1 Add i TESEr) to ae. Ans. Ee ae 
1 1 2 
2. Add 1. z to ral Be Ans. Remy 
‘a ted a—b 4ab 
3 From A 7 ai ca Ans. ass 
% *. 
1+ 2? bal et 4x? 
4 ow mee Bie aipraret Ans. Tagore 
ee x 9x + 20 x? — 13x + 42 
5. EM 7} 6x by ae 
VE ae 
aan x lla + 28 
x2 
: xt — bt x? + be 
ee he (MEER 2 
6. Multiply 24 ibe Le by Pigs. Ans. «3 + bx. 
Wa DIVIde >, (eter ot, ih 
a—x a+e2r a—x ate 
j H a? + x2 
Ss. 7 ae 
ie n—l n—l : 
8. Divide } fe PEC by 1— gE UET Ans. n. 


CHAP. 1V,] EQUATIONS OF THE FIRST DEGREE. 59 


: 


/ 


CHAPTER IV. 
OF EQUATIONS OF THE FIRST DEGREE. 


79. An Equation is the algebraic expression of two equal quan- 
ties with the sign of equality placed between them. Thus, 


a=-a+t ob 


: alee ; sty 
is an equation, in which @ is equal to the sum of a and 6. 


80. By the definition, every equation is composed of two parts, 
separated from each other by the sign =. ‘The part on the 


left of the sign, is called the first member, and the part on the 


right, is called the second member; and each member may be 
composed of one or more terms. 


81. Every equation may be regarded as the enunciation, in al- 
gebraic language, of a particular problem. Thus, the equation 


ea 30, 


is the algebraic enunciation of the following problem: 


| 


| 





To find a number which, being added to itself, shall give a sum 


equal to 30. 


Were it required to solve this problem, we should first express 
it in algebraic language, which would give the equation 


| e- .g@ —' 30, 
by adding « to itself, - - 2x = 30, 
and dividing by 2, - - x = 15. 


Hence we see that the solution of a problem by algebra, con- 
sists of two distinct parts: viz., the statement, and the solution of 


an equation. 


60 ELEMENTS OF ALGEBRA. [CHAP. IV. 


The sTATEMENT consists in finding an equation which shall ex 
press the relation between the known and unknown quantities of the 
problem. 

The soturion of the equation consists in finding such a value 
for the unknown quantity as being substituted for it in the equa- 
tion will satisfy it; that is, make the first member equal to the 
second. 


82. An equation is said to be verified, when such a value is 
substituted for the unknown quantity as will prove the two mem- 
bers of the equation to be equal to each other. 


83. Equations are divided into classes, with reference to the 
highest exponent with which the unknown quantity is affected. 

An equation which contains only the first power of the un- 
known quantity, is called an equation of the first degree: and 
generally, the degree of an equation is determined by the greatest 
of the exponents with which the unknown quantity is affected, 
without reference to other terms which may contain the unknown 
quantity raised to a less’ power. Thus, 


ae + 6 =cex+d is an equation of the 1st degree. 
Qa? — 34 =5 — 2x? is an equation of the 2d degree. 
4x3 — 5x? = 3u +11 is an equation of the 3d degree. 


if more than one unknown quantity enters into an equation, its 
degree is determined by the greatest sum of the exponents with 
which the unknown quantities are affected in any of its terms. 
Thus, 
ry + ber =d‘ is of the second degree. 
xyz? -+ cx? — a> is of the fourth degree. 


84. Equations are also distinguished as numerical equations and 
literal equations. ‘The first are those which contain numbers only, 
with the exception of the unknown quantity, which is always de- 
noted by a letter. Thus, 

4e—3=—2r+ 5, 322?—av= 8, 
are numerical equations. They are the algebraical translation of 
problems, in which the known quantities are particular numbers. 

A literal equation is one in which a part, or all of the known 
quantities, are represented by letters. Thus, 

ba? + az — 32 = 5, and cr+ dx? =e+f, 
are literal equations. ; 


‘CHAP. IV.] EQUATIONS OF THE FIRST DEGREE. 61 
i 85. It frequently occurs in Algebra, that the algebraic sign + 
‘or —, which is written, is not the true sign of the term before — 
which it is placed. Thus, if it were required to subtract —b 
‘from a, we should write ; P 


a—(—b)=a+b. . 


Here the true sign of the second term of the binomial is plus, 
although its algebraic. sign, which is written in the first member 
of the equation, is —. This minus sign, operating upon the sign 
‘of b, which is also negative, produces a plus sign for 4 in the 
result. The sign which results, after combining the algebraic 
‘sign with the sign of the quantity, zs called the essential sign of 
the term, and is often different from the algebraic sign. 
By considering the nature of an equation, we perceive that it 
‘must possess the three following properties: 
' 1st. The two members are composed of quantities of the same 
kind. 

2d. The*two members are equal to each other. 
_ 3d. The essential sign of the two members must be the same. 


86. An axiom is a self-evident proposition. We may here state 
the following : 
1. If equal quantities be added to both members of an equa- 
‘tion, the equality of the members will not be destroyed. 
/ 2 If equal quantities be subtracted from both members of an 
‘equation, the equality will not be destroyed. 
3. If both members of an equation be multiplied by the same 
jnumber, the equality will not be destroyed. 

4. If both members of an equation be divided by the same 
number, the equality will not be destroyed. 








~ 


Solution of Equations of the Furst Degree. 


87. The transformation of an equation is any operation by 
\which we change the form of the equation without affecting the 


Furst Transformation. 


88. When some of the terms of an equation are fractional, to 
|reduce the equation to one in which the terms shall be entire. 


62 ELEMENTS OF ALGEBRA. (CHAP. IV. 


Take the equation, 
201 3 x 


First, reduce all the fractions to the same denominator, by the 
known rule; the equation then becomes 


48x 54a 120 
Recipe hig aaa 

If now, both members of this equation be multiplied by 72, the 
equality of the members will be preserved, and the common de- 
nominator will disappear; and we shall have 


48x — 54x + 124 = 792; 
or dividing by 6, Sa Ox He Qe ee. 


89. The last equation could have been found in another man- 
ner by employing the least common multiple of the denominators. 

The common multiple of two or more numbers ‘is any number 
which each will divide without a remainder; and the least com- 
mon multiple, is the least number which can be so divided. 

The least common multiple can generally be found by inspec- 
tion. Thus, 24 is the least common multiple of 4, 6, and 8; and 
12 is the least common multiple of 3, 4, and 6. 

Take the last equation, 


We see that 12 is the least common multiple of the denomina- 
tors, and if we multiply each term of the equation by 12, divi- 
ding at the same time by the denominators, we obtain 

8x — 9x + 2x = 132, 
the same equation as before found. 


90. Hence, to transform an equation involving fractional terms 
to one involving only entire terms, we have the following 


RULE. 


‘Form the least common multiple of all the denominators, and then 
multiply every term of the equation by it, reducing at the same 
time the fractional to entire terms. 


|) 


| 
H 


t 
} 
i 


j 
CHAP. IV.] EQUATIONS OF THE FIRST DEGREE. 


: EXAMPLES. FU 


| 1. Reduce = + = —3= 20, to an equation involving entire 


terms. bd 
_ We see, at once, that the least common multiple is 20, by 
which each term of the equation is to be multiplied. 





x 20. 
| Now, RereeaO tes) Moe siemi4d, 
2 20 


chat is, we reduce the fractional to entire terms, by multoplyng 
the numerator by the quotient of the common multiple divided by 
‘the denominator, and omitting the denominators. 


Hence, the transformed equation is 
: 


4x¢ + 5x — 60 = 400. 
x 
oy 


' 2. Reduce 
‘entire terms. Ans. 7x + 52 —140 = 105. 


a 8 


—4=3 to an equation involving only 


_ 3. Reduce = -- = +f=g to an equation involving only en- 


vite terms. Ans. ad — be + bdf = bdg. 
4, Reduce the equation 
an cx toa aa? 5 Qc* 
MM oh aks* yas 
0 one involving only entire terms. 


Ans. atbx — 2a%be?x + 4atb? — 4b3c2x — 5a8 + 2a2h2c? — 34353, 


~ 





: 


Second Transformation. 


91. When the two members of an equation are entire polyno- 
mials to transpose certain terms from one member to the other. 
Take for example the equation 5x—6—=8+2z. 

If, in the first place we subtract 
tx from both members, the equality 5x —6 — 2x = 8+ 2x7 —2y: 
vill not be destroyed, and we have 
wr, by reducing the terms in the 


second member, ‘ S52 — 6 — 2x = 8. 


64 ELEMENTS OF ALGEBRA [CHAP. IV. 


Whenbe. we_see th at the term 2x2, which was’ additive in the 
— 


second member, becomes subtractive in the first. 
. In the second place, if we add 6 
to both members, the equality will 5a4—6 —2x+6=8+ 6; 
still exist, and we have 
or, since —6 and +6 destroy each other Sx —2r27=8-+ 6. 
Hence, the term which was subtractive in the first member, 





passes into the second member with the sign plus. | 
For a second example, take the equation 
ax +b=—d— cz. 
If we add cx to both 
members and subtract 6, >) ax +6b+cr—b=d—ca+ cr —b); 
the equation becomes 
or reducing - - - ax+ecx=d—b. 


Hence, we have the following principle: 


ie Any term of an equation may be transposed from one member 
to the other by changing its sign. 


92. We will now apply the preceding principles to the resolu- 
tion of equations. 


1. Take the equation 4% —3 = 2x+ 5. 
By transposing the terms — 3 and 2a, it becomes 


4g — 9a, 5 ae 


and by reducing ties 8: 
dividing by 2 x =; =e . * 


Now, if 4 be substituted in the place of # in the given equa- 
tion, it becomes 
A idee 3 — 2. id ae 
that is, BOs 41 3 


Hence, 4 is the true value of «; for, being substituted spot: x in 
the given equation, that equation is verified. : 


2. For a second example, take the equation 


ron ee 7 
2 Se 


12 #3 S | 6a a 





. CHAP. IV.] EQUATIONS OF THE FIRST DEGREE. 65 4 


By making the denominators disappear, we have 


10x — 32% — 312 = 21 — 52a 
by transposing 102 — 32% + 52x = 21 + 312 


by reducing 302 = 333 

333s 111- 
wid b — Se Sadie wl: 
dividing by 30 Bree ah il gl 


_a result which, being substituted for «, will verify the giver 
equation. 


: 


3. For a third example let us take the equation 

(3a — x) (a — b) + 2ax = 4b (a + a). 
It is first necessary to perform the multiplications indicated, in 
‘order to reduce the two members to polynomials, and thus be 
able to disengage the unknown quantity « from the knawn quan- 
tities. Having performed the multiplications, the equation he- 
comes, 
3a? — ax — 3ab+ br + 2ax = 4be + 4ab; 
by transposing — ax + bx + 2ax — 4bx = 4ab + 3ab — 3a’, 





by reducing ax — 3bx = Tab — 3a?; 
or, (Art. 48), (a — 3b)a = Tab — 3a?. 
Dividing both members by a — 3d, we find 
__ Tab — 3a? 
—  a—3b 


93. Hence, in order to resolve any equation of the first de- 
gree, we have the following general 


RULE. 


I. If the equation contains fractional terms, reduce it to one in 
| which all the terms shall be entire, and then transpose all the terms 
affected with the unknown quantity into the first member, and ail 
| the known terms into the second. 

II. Reduce to a single term all the terms involving the unknown 
quantity: this term will be composed of two factors, one of which 
will be the unknown quantity, and the other all tts co-efficients con- 
nected by their respective signs. 

Ill. Then divide both members of the equation by the multiplier 


of the unknown quantity. 7 





\65 
; 


10. 


ll. 


12. 


13. 


. Given 


. Given 


. Given 


. Given 


. Given 


. Given 


. Given 


. Given 


Given 


Given 


Given 


Given 


ELEMENTS OF ALGEBRA. [CHAP. Tv. 
EXAMPLES. 
3a — 2 + 24 = 31 to find a. Ans. x« = 3. 
e+ 18 = 3x — 5 to find z. Ans? ees tre, 
6 — 22,+ 10 = 20 — 3x2 — 2 to find z. 
Ans. ¢ = 2. 


1 1 
tp oe to Lt to find a. Ans. x = 6. 





























2x — a t1—5e—2 to find a. Ans. © =~ 
San + — —3 = be—a to find a. 
6. — 3a 
Ans. hee ae OT yh 
x—3 x x 1 
5) as chain to find a. 
Ans. & = 234. 
a+3 Hy x—5 
— — 4 — ; 
5 tt ' to find « 2 | 
Ans. 2 = 3%. 
ax — b a ba bx — a 
ty i li a t 
% Dis 5 5 o find x 
3d . 
A —_ J 
me 3a —. 25 
Sax Qha 
Se ee 
; 7 f, to find x 
cdf + 4cd 
Ans. «= 
> Gea 
8ax — b 20 ae 
ue —4— °, 
7 5 b, to find x 
Ans. Ppiiipddla 200 abc 
loa 
x x— 2 x 13 
—_—— ——==--~~, t 2 
P - as ne o find 
Ans. « = 10. 
x x x x 
ab Se. @ 
Ans. x = any" 


ved — acd + adb — abe’ 





‘CHAP 1V,] EQUATIONS OF THE FIRST DEGRER. 67 


oa — 5 Ag == 2 




















14. Given «2 — 13 + 7 = +1, to find x. 
Ans. x = 6. 
“ne x 8x. «—3 
15. Given Tg TE = + 1228, to find a. 
Ans, x == 14, 
‘ —2 3x2 — 
16. Given 2x2 — i heb a" a to find x. 
5 2 > 
Ans. x = 3 
—d " 
17. Given 3x2 + ae =«e-+a, to find z. 
And. 0 se + 
18. Find the value of x in the equation 
(a + b)(a —b) x 4ab — B? a*® — br 
ne Re aa OM Dae 
a* + 3a%b + 4a7b? — 6ab3 +. 254 
Ans: x = a 





26 (2a? + ab — b?) : 


Questions producing Equations of the First Degree, involving 
but one Unknown Quantity. 


94. It has already been observed (Art. 81), that the solution 
f a problem by Algebra, consists of two distinct parts. 

ist. The statement; and 

2d. The solution of the equation. 

We have already explained the methods of solving the equa- 
on; and it only remains to point out the best manner of making 
le statement. 

This part cannot, like the second, be subjected to any- well- 
efined rule. Sometimes the enunciation of the problem furnishes 
1e equation immediately; and sometimes it is necessary to dis- 
jover, from the enunciation, new conditions from which an equa- 
on may be formed. | | 
The conditions enunciated are called explicit conditions, and 
lose which are deduced from them, implicit conditions. 

*In almost all cases, however, we are enabled to discover the 
uation by applying the following 


6S ELEMENTS OF ALGEBRA. (CHAP. IV. 


RULE. 


Represent the unknown quantity by one of the final letters of the 
alphabet, and then indicate, by means of the algebraic signs, the 
same operations on the known and unknown quantities, as would 
verify the value of the unknown quantity, were such value known. 


QUESTIONS. 


1.Find a number such, that the sum of one half, one third 
and one fourth of it, augmented by 45, shall be equal to 448. 
Let the required number be denoted by - - - - @. 


Then, one half of it will be denoted by - - - = 


one third of it - - - - by- - 


é 


5 
] 
mila wle wie 


one fourth of it- - - - by- = - 
And by the conditions, Se + = + on + 45 = 448. 
Now, by subtracting 45 from both members, 
x xz x 
3 emer uaaly vss 


By making the terms of the equation entire, we obtain 
6a + 4a + 32 = 4836; 


or - - 132 = 4836. 
4836 
Hence - z= ar? — 372. 


Let us see if this value will verify the equation of the prob- 
lem. We have 


Se SE + 45 = 186 + 124 + 93 + 45 = 448, 


2. What number is that whose third part exceeds its fourth, 
by 16. 


Let the required number be represented by 2. Then 


a2 — the third part. 


2 = the fourth part. 


CHAP. IV.) EQUATIONS OF THE FIRST DEGREE. 69 


And by the question - 2 — : em 10 
or, - = = + 4H — 37 = 192. 
wi=i192: 
Verification. - 
{ 192 sada ree lay: 
1c: galiaaa Mibiladaiec aii Gil ae 


3. Out of a cask of wine which had leaked away a third part, 
21 gallons were afterward drawn, and the cask being then 
gauged, appeared to be half full: how much did it hold? 


| Suppose the cask to have held 2 gallons. 


Then, = = what leaked away. 
And = + 21= what leaked out, and what was drawn 
x 1 ‘ 
Hence, 7 + 21= ria by the question. 
or 22 + 126 = 32. 
or — x = — 126. 
or fee 126, 


by changing the signs of both members, which does not destroy 
their equality. 
Verification. 
+21 = 42 +4 21 = = 68. 


4. A fish was caught whose tail weighed 9/b.; his head weighed 
as much as his tail and half his body, and his body weighed 
as much as his head and tail together: what was the weight of 
the fish ? 

bet - = 22 = the weight of the body. 

Then - -9+2a= weight of the head. 

And since the body weighed as much as both head and tail 
227=-9+ 9+ 
Oris §6 = =! 2a —e¢= 18 


and en BN x — 18. 


70 ELEMENTS OF ALGEBRA. [CHAP. IV 


Verification. 


22 = 36/) = weight of the body. 
9+2—=27lb= weight of the head. 

9/5 = weight of the tail. 

Hence, 7216 = weight of the fish. 


5. A person engaged a workman for 48 days. For each day 
that he labored Jhe received 24 cents, and for each day that he 
was idle, he paid 12 cents for his board. At the end of the 48 
days, the account was settled, when the laborer received 504 
cents. Required the number of working days, and the number of 
‘days he was idle. 

If these two numbers were known, by multiplying them respec- 
tively by 24 and 12, then subtracting the last product from the 
first, the result would be 504. Let us indicate these operations 
by means of algebraic signs. 


Let - - a = the number of working days. 
Then 48—x = the number of idle days. 
24 xX « = the amount earned, and 
12(48 — x)= the amount paid for his board. 
Then 24a —12(48—2) = 504 what he received. 
or 242 —576+ 122 = 504. | 
or 36a = 504 + 576 — 1080 
and = fe = 30 the working days. 
whence, 48 —30=18 the idle days. 
Verification. 
Thirty day’s labor, at 24 cents a day, 
amounts to a fie OBIT AH a et BO Seon Cte, 
And 18 days’ board, at 12 cents a day, 
amounts to et Reb ne = a 


And the amount received is their difference : 504. 








CHAP. IV.] EQUATIONS OF THE FIRST DEGREE. O71 


General Solution. 


n== the whole number of working and idle days, 
a= the amount received for each day he worked, 
b= the amount paid for his board, for each idle day, 
c = the balance due, or the result of the account. 
# = the number of working days, 

v= the number of idle days. 
Then, ax — what he earned; 
and, b(n—«)= the amount deducted for board. 


The equation of the problem will then be, 


ax —b(n—az) =e 








whence \ ac— bn +ba=ce 
(a+b)ex—c +bn 
c + bn 
, aha TTD 
c +bn an-+bn—c—bn 
and consequently, er ek 7. 7 
an—c 
or 2 — 2¢'=S etek 


6. A fox, pursued by a greyhound, has a start of 60 leaps. 
He makes 9 leaps while the greyhound makes but 6; but 3 
leaps of the greyhound are equivalent to 7 of the fox. How 
many leaps must the greyhound make to overtake the fox? 

From the enunciation, it is evident that the distance to be 
passed over by the greyhound, is equal to the 60 leaps of the 
fox, plus the distance which the fox runs after the greyhound 
starts in pursuit. 

Let «= the number of leaps made by the greyhound from 
the time of starting till he overtakes the fox. 

Now, since the fox makes 9 leaps while the greyhound makes 
6, the fox will make 14, or 2 leaps while the greyhound 
makes 1; and, therefore, while the greyhound makes «a leaps, the 


3 
fox will make ths leaps. Hence, 


3 


72 ELEMENTS OF ALGEBRA. (CHAP. IV 


the number of leaps made by the fox, in passing over the entire 
distance. 

It might, at first. be supposed that the equation of the problem 
would be obtained by placing this number equal to a; but in. 
doing so, a manifest error would be committed; for the leaps of 
the greyhound are greater than those of the fox, and we should 
thus equate numbers referred to different units. Hence, it is ne- 
cessary to express the leaps of the fox by means of those of 
the greyhound, or reciprocally. 

Now, according to the enunciation, 3 leaps of the greyhound 
are e,‘ivalent to 7 leaps of the fox; and hence, 1 leap of the 


OP 7 
greyhound -s equivalent to > leaps of the fox; consequently, 


x leaps of the gre,'-sund are equivalent -to & of the fox: that 
is, had the leaps of the gre. “ound been no longer than those of 


rm k 
the fox, he would have made g bere instead of a leaps. 


bina j 7 
Hence the true equation is, ; = 60 +2; 


or, by making the terms entitfe 14% = 360 + Qa, : 
whence, = -) = eb re hme 360 . anid are: 


Therefore, the greyhound will make 72 leaps to overtake the fox, 
’ 


and during this time the fox will make 72 x <= 108. 
Verification. 


The 72 leaps of the greyhound are equivalent to 
72 X 7 





= 168 leaps of the fox = the whole distance. 
And 60+ 108 = 168, the leaps which the fox made from the 
beginning. 


7. A can do a piece of work alone in 10 days, and B in 13 
days: in what time can they do it if they work together ? 
Denote the time by a, and the work to be done by 1. Then 


in 1 day A could do 5 of the work, and B could do ro of 










| CHAP. IV.] EQUATIONS OF THE FIRST DEGREE. 73 


‘it; and in « days A could do - of the work, and B =: 


‘hence, by the conditions of the question, 





io igi 

which gives 13x + 10x = 130: 

130 * 

| bence, 28m 371 30.01% = Sia 543 days. 


’ 


| 8. Divide $1000 between A, B, and C, so that A shall have 
$72 more than B, and C $100 more than A. 
i Ans. A’s share = $324, B’s = $252, C’s = $424. 


i 


9. A and B play together at cards. A sits down with $84 
‘and B with $48. Each loses and wins in turn, when it ap- 
/pears that A has five times as much as B. How much did A 


win ? Ans. $26. 


_ 10. A person dying leaves half of his property to his wife, one 
‘sixth to each of two daughters, one twelfth to a servant, and the 
iremaining $600 to the poor: what was the amount of his prop- 
lerty? Ans. $7200. 


| 11. A father leaves his property, amounting to $2520, to four 
isons, A, B, C, and D. C is to have $360, B as much as C 
‘and. D together, and A twice as much as B less $1000: how 
‘much does A, B, and D, receive? 

| Ans. A $760, B $880, D $520. 


12. An estate of $7500 is ‘to be divided between a widow, two 
sons, and three daughters, so that each son shall receive twice as 
‘much as each daughter, and the widow herself $500 more than 
all the children: what was her share, and what the share of 
jeach child? 3 Widow’s share $4000. 
Ans Bea son $1000. 
Each daughter $500. 





13. A company of 180 persons consists of men, women, and 
children. ‘The men are 8 more in number than the women, and 
the children 20 more than the men and women together: how 
many of each sort in the company ? 

Ans. 44 men, 36 women, 100 children. 


74 ELEMENTS OF ALGEBRA. (CHAP. IV 


14. A father divides $2000 among five sons, so that each elde: 
should receive $40 more than his next younger brother: what i: 
the share of the youngest ? Ans. $320. 


15. A purse of $2850 is to be divided among three persons 
A, B, and C; A’s share is to be to B’s as 6 to 11, and C i 
to have $300 more than A and B together: what is each one’ 
share ? Ans. A’s $450, B’s $825, C’s $1575. 


16. Two pedestrians start from the, same point; the first step: 
twice as far as the second, but the second makes 5 steps whil 
the first makes but one. At the end of a certain time they ar 
300 feet apart. Now, allowing each of the longer paces to be : 
feet, how far will each have travelled? 

Ans. 1st, 200 feet; 2d, 500. 


17. Two carpenters, 24 journeymen, and 8 apprentices, re 
ceived at the end of ‘a certain time $144. The carpenter 
received $1 per day, each jourueyman half a dollar, and eae! 
apprentice 25 cents: how many days were oct employed ? 

Ans. 9 days. 


18. A capitalist receives a yearly income of $2940: four fifth 
of his money bears an interest of 4 per cent., and the remainde 
of 5 per cent.: how much has he at interest? Ans. $70000. 


19. A cistern containin; 60 gallons of water has three unequa 
cocks for discharging 1; the largest will empty it in one how 
the second in two hours, and the third in three: in what tim 
will the cistern be emptied :f they all run together ? 

Ans. 3238, min. 


20. In a certain orchard 4 are apple-trees, 1 peach-trees 
4 plum-trees, .20 cherry-trees, and 80 pear-trees: how man 


trees in the orchard ? Aas, eave 


21. A farmer being asx~! how many sheep he had, answere 
that he had them in hve fields; in the 1st he had 4, in th 
2d 3, in the 3d 4, in the 4th rz, and in the 5th 450: ho 
many had he? ) Ans. 1200. 

22. My horse and saddle together are worth $132, and th 


horse is worth ten times as much as the saddle: what is th 
value of the horse ? Ans. $120. 


| OHAP. IV.] EQUATIONS OF THE FIRST DEGREE. v5) 


_ 23. The rent of an estate is this year 8 per cent. greater than 
‘it was last. This year it is $1890: what was it last year? 


Ans. $1750. 
24, What number is that from which, if 5 be subtracted, 2 of 
the remainder will be 40? Ans. 65. 


25. A post is + in the mud, 4 in the water, and ten feet above 
the water: what is the whole length of the post? 
| Ans. 24 feet. 


_ 26. After paying 4 and } of my money, I had 66 guineas left 
‘In my purse: how many guineas were in it at first? 
! Ans. 120. 


27. A person was desirous of giving 3 pence apiece to some 
‘beggars, but found he had not money enough in his pocket by 8 
pence; he therefore gave them each two pence and had 3 pence 


_remaining: required the number of beggars. Ans. 
i 


28. A person in play lost + of his money, and then won 3 
‘shillings ; after which he lost + of what he then had; and this 
| done, found that he had but 12 shillings remaining: what had 
he at first ? Ans. 20s. 


/ 29. Two persons, A and B, lay out equal sums of money in 
trade; A gains $126, and B loses $87, and A’s money is now 
double of B’s: what did each lay out? , Ans. $300. 





, 30. A farmer bought a basket of eggs, and offered them at 7 
,cents a dozen. But before he sold any, 5 dozen were broken 
by a careless boy, for which he was paid. He then sold the re- 
/mainder at 8 cents a dozen, and received as much as he would 
have got for the whole at the first price. How many eggs had 
he in his basket 2 Ans. 40 dozen. 


31. A person goes to a tavern with a certain sum of money in 
his pocket, where he spends 2 shillings; he then borrows as 
.much money as he had left, and going to another tavern, he 
_there spends 2 shillings also; then borrowing again as much 
| money as was left, he went to a third tavern, where likewise 
| he spent 2 shillings and borrowed as much as he had left; and 
| again spending 2 shillings at a fourth tavern, he then had nothing 
}remaining. What had he at first? 





Ans. 3s. 9d ~ 


76 ELEMENTS OF ALGEBRA. (CHAP. IV. 


Of Equations of the First Degree, involving two or more 
Unknown Quantities. 

95. Although several of the previous questions contained in their 
enunciation more than one unknown quantity, we have neverthe- 
less resolved them all by employing but one symbol. The rea- 
son of this is, that we have been able, from the conditions of the 
enunciation, to represent the other unknown quantities by means 
of this symbol and known quantities; but this cannot be done 
in all problems containing more than one- unknown quantity. 

To explain the methods of resolving problems of this kind, let 
us take some of those which have been resolved by means of 
one unknown quantity. 


1. Given the sum of two numbers equal to a, and their differ 
ence equal to ); it is required to find the numbers. 


Let x= the greater, and y the less number. 
Then by the conditions at+y=a; 
and e—y=b, 


By adding (Art. 86, Ax. 1), 2a =a+b. 
By subtracting (Art. 86, Ax. 2), 2y=a—b. 


Each of these equations contains but one unknown quantity 











b 
From the first we obtain a ms : 
a— Dp 
And from the second Yom ay 
Verification. 
ety ae and 6 ee eee 
2 2 2 2 2 2 


2. A person engaged a workman a number of days, denoted 
by x. For each day that he labored he was to receive a cents, 
and for each day that he was idle he was to pay 6b cents for his 
board. At the end of the n days, the account was settled, when 
the laborer received ¢ cents. Required the number of working 
days and the number of days he was idle. 


Let x2 == the number of working days. 
== the number of idle days. 
Then, ax = what he earned, 


and by = what he paid for his board ; 


IHAP, IV.] EQUATIONS OF THE FIRST DEGREE. 17 


e-iy=n 
ax — by =c. 

It has already been shown that the two members of an equa- 
jon can be multiplied by the same number, without destroying 
‘he equality; therefore, multiply both members of the first equa- 
ion by 6, the co-efficient of y in the second, and we have 


; and by the question, we have 


he equation- - - - - dba+by=din, 

which, added to the second - axr—by=c, 

rives Sa Se ee ne |) aa bas bn +b. ce, 
il bn +e 





Whence- - - - - 


ry a tend 

_ In like manner, multiplying the two members of the first equa- 
lon by a, the co-efficient of x in the second, it becomes 

| ax + ay = an; 

tom which, subtract the second equation, ax — by = ec, 





ISS a ae ay + by = an —c. 
Wh tare gree” Gtr ten). ra an —cC 
ence y Rat 


By introducing a symbol to represent each of the unknown 
quantities of the problem, the above solution has the advantage 
of making known the two required numbers, independently of each 
other. 


What will be the numerical values of x and y, if we suppose 


n=48, a= 24,-6=12, and c= 504. 


Elimination. 
| 96. The method which has just been explained, of combining 
;wo equations, involving two unknown quantities, and deducing 
herefrom a single equation involving but one, may be extended 
1 three, four, or any number of equations, and is called Elimna- 
ton. 
There are three principal methods of elimination: 
_ Ist. By addition and subtraction. 
_ 2d. By substitution. 
3d. By comparison. 
We shall discuss these methods separately. 





78 ELEMENTS OF ALGEBRA. [CHAP. IV 


Elimination by Addition and Subtraction. 


97. Before considering the case of Elimination, we will ex-— 


plain a new notation which is about to be used. 
It often happens, in Algebra, that some of the known quantities 


of an equation or problem, though entirely independent of each 


other in regard to their values, have, nevertheless, certain rela- 
tions which it is desirable to preserve in the discussion. In such 
case, the second quantity is represented by the same letter, with 
a small mark over it. Thus, if the first quantity was denoted by 
a, the second would be denoted by a’, and is read, a prime, 
If there were a third, it would be denoted by a”, and read, 
a second, &c. 


Let us now take the two equations, 
ax -- by =e 
an by =e". 
If the co-efficients of either of the unknown quantities were the 
same in both equations; that is, if @ were equal to a’, or 0b 
to &’, we might by a simple subtraction form a new equation that 


would contain but one unknown quantity; and from this equa- 
tion, the value of that unknown quantity could be deduced. 


If, now, both members of the first equation be multiplied by J’, 


the co-efficient of y in the second, and the two members of the 
second by 6, the co-efficient of y in the first, we shall obtain 
ab’x + bb’y = b’e 
abe + bb’y = be’; 
and by subtracting the second from the first, 
(ab’ — a/b) x= l’c — be’; 
whence anit Oh 
: serie} 


If we multiply the first of the given equations by a’, and the 
second by a, we shall have 


aad’x + a’by = a’e 
aan + ab’y = atc’; 


and by subtracting the second from the first, 








HAP. Iv.]. EQUATIONS OF THE FIRST DEGREE. ~ 79 


(ab — ab’) y= ac — ac’; 

_ &e— ac 

is oer 
rt, if we wish the value for y to have the same denominator 
‘jth that for #, we change the signs of the numerator and de- 
Jominator, and write 


vhénce, 





ac’ —a’e 
ae ee: 
| The method of elimination just explained, is called the methoa 
y addition and subtraction, because the unknown quantities dis- 
ppear by additions and subtractions, after having prepared the 
quations in such a manner that the same unknown quantity shall 
‘ave the same co-efficient in both equations. 
b’c — bec’ ad —ae 
al —ab’ *~ a —ab 
educed from the equations 
| ax + by =c 
/ vet vy=c’ 


The formulas r= 


rill enable us to write the values of x and y immediately, with- 
ut the trouble of elimination. ‘They contain the germe of a gen- 
ral rule, not before given, for the solution of all similar equations. 


RULE. il 


I. The first term mn the numerator for the value of x, is found 
y beginning at b’ and crossing up to c—giving b’c; the second 
: is found by crossing from b to c’—giving be’. 

Il. For the first term in the numerator of the value for y, begin 
t a and cross down to ¢e/—giving ac’; and for the second term, 
ross from a! to c—giving a’c. 

Ill. The first term of the common denominator is found by cross- 
ag from a to b’—giving ab’; and the second, by crossing from 
to b—giving ab. 

' The manner of obtaining these formulas will be easily remem- 
. and their applications will be found very simple. 

1. What are the values of x and y in the equations, 


ox) -++ Ty = 43 
llx + 9y = 69. 


80 ELEMENTS OF ALGEBRA. [CHAP. 1V. 


We write immediately, 


























9x43— 7x69 387 — 483 — 96 ; 
-=5 x 9 ipl Mn 7. ao ee ee 
25 X69 to) St ae eee 
* — 32 ie — 3265) 9 NB20T o- 
2. What are the values of 2 and y in the equations, 
y 
oF Te 
2—4y => —11. 
We write 
—4x14—(—}x-—11)_ —56-—Y 
TTS S42 ge eee ee ae 
3x SiS 
Si —12+414 ——124+1 7 " 


Elimination by Substitution. 
98. Let us take the two equations 
or + Ty = 43 and lla + 9y = 69. 
Find the value of « in the first equation, which gives 


Substitute this value of « in the second equation, and we have 


43 —.7 
11 x T+ + oy = 69. 


or 473 — 77y + 45y — 345: 
or — 32y = — 128. 
Hence y= 4. 
43 — 28 
And hen a i 


This method, called the method by substitution, consists in find- 
ing in one equation the value of one of the unknown quantities, 
as if the others were already determined, and then substituting 
this value in the other equations. In this way, new equations 
are formed from which one of the unknown quantities has been 
eliminated. We then operate in a similar manner, on the new 
equations. 


CHAP. IV.] EQUATIONS OF THE FIRST DEGREE. 81 


Elimination by Comparison. 
99. Let us take the two equations, 
5a + 7y = 43 and llx + 9y = 69. 


Finding the value of x in the first equation, we have 


43 —7 
a= ae 
) 5 
And finding the value of « in the second, we obtain 
69 —9 
Pp cnia 
Dy 


Let these two values of « be placed equal to each other, and 


. se ty 60 Oy 


we have, 5 poets Ga 5 
or, 473 — T7y = 345 — 45y; 
or, — 32y = — 128. 

| Hence, y ezci4 

69 — 36 
and, t = oraque 


_ This method of elimination is called the method by compari- 
‘son, and consists in finding the value of the same unknown 
quantity in all the equations, and then placing those values equal 
to each other, two and two. This will give rise to a new set of 
equations containing one less unknown quantity,.and upon which 
We operate as on the given equations. 

The new equations which arise, in the last two methods of 
elimination, contain fractional terms. This inconvenience is avoid- 
ed in the first method. The method by substitution is, however, 
advantageously employed whenever the co-efficient of either of 
the unknown quantities in one of the equations is equal to unity, 
because then the inconvenience of which we have just spoken 
does not occur. We shall sometimes have occasion to employ 
this method, but_generally the method by addition and subtraction 
‘is preferable. When the co-efficients are not too great, we can 
‘perform the addition or subtraction at the same time with the 
Multiplication which is made to render the co-efficients of the 
“same unknown quantity equal to each other. 


~ 


82 ELEMENTS OF ALGEBRA. (CHAP. IV, 


100. Let us now consider the case of three equations involving 

three unknown quantities. 
or — Oy + 42 = 15. 
Take the equations, 7x + 4y — 3z2= 19. 
22+ y + 62— 46, 

To eliminate z from the first two equations, multiply the first 
equation by 3 and the second by 4; and since the co-efficients 
of z have contrary signs, add the two results together: this gives 
a new equation - - - - - 43x” — 2y = 121 

Multiplying the second equation by 2, a fac- 
tor of the co-efficient of z in the third equa- 
tion, and adding them together, we have 162 + 9y = 84 

The question is then reduced to finding the values of x and Y; 
which will satisfy these new equations. 

Now, if the first be multiplied by 9, the second by 2, and the 
results be added together, we find 

419x = 1257, whence #—3. 


By means of the two equations involving x and y, we may de- 
termine y as we have determined x; but the value of y may be 
determined more simply, by observing, that by substituting for a 
its value found above, the last of the two equations becomes, 


48 + 9y = 84, whence y= ———— —4, 


In the same manner, by substituting the values of x and y, the 
first of the three proposed equations becomes, 


24 


15 — 24+ 42=15, whence z= 7 = G6. 


101. Hence, if there are m equations involving a like number 
of unknown quantities, the unknown quantities may be eliminated 


. by the following 


RULE. 


I. To eliminate one of the unknown quantities, combine any one 
of the equations with each of the m—1 others; there will thus 
be obtained m—1 new equations containing m—1 unknown quan- 
tities. 

Il. Eliminate another unknown quantity by combining one of these 
new equations with the m — 2 others; this wil give m—2 equa 
tions containing m— 2 unknown quantities. 





HAP. IV.] EQUATIONS OF THE FIRST DEGREE. 83 












III. Continue this series of operations until a single equation is 
“tained containing but one unknown quantity, the value of which 
.an then be found. Then by going back through the series of equa- 


ions the values of the other unknown quantities may be successively 
etermined. 


102. It often happens that some of the proposed equations do 
‘ot contain all the unknown quantities. In. this case, with a 
ttle address, the elimination is very quickly performed. 


Take the four equations involving four unknown quantities, 


| 2e — By + 22 — 13 - - (1) 4y4+2z2=14 - .- (3). 
rect es SOE) OY OU SS aes (a 


1 By examining these equations, we see that the elimination of 
In equations (1) and (3), will give an equation involving = 
ad y; and if we eliminate u in the equations (2) and (4), we 
1all obtain a second equation, involving x and y. In the first 
ace, the elimination of z, in (1) and (3) gives 7y—2xe— 1 


lat of w, in (2) and (4), gives. - - - 20y + 6x = 38 
' Multiplying the first of these equations by 

_and and adding, We have- - -~ .« 4ly = 41 
whence - - - - ” - - eh mt 
| Substituting this value in Ty —2x=1, we 

id - a ~ ~ - - - Eo plat 
“Substituting for « its value in equation (2), 
| becomes, 4u— 6 — 30, whence . - us 9 
And substituting for y its value in equation 

), there results - - - - “ - a= 5 

\ 


. 


Of indeterminate Problems. 


103. In all the preceding reasoning, we have supposed the 
,mber of equations equal to the number of unknown quantities. 
‘us must be the case in every problem, in order that it may be 
terminate; that is, in order that it may admit of a finite num- 
it of solutions. 


,Let it be required, for example, to find two quantities such, 
»t five times one of them, diminished by three times the other, 
rall be equal to 12. 


.. » there results, «= 3, 


84 ELEMENTS OF ALGEBRA. (CHAP. IV. 


If we denote the quantities sought by 2 and y, we shall have 
the equation 


5a — 3y = 12, 
+e oe 
= 2a 


whence, 


Now, by making successively, 
pice 1). 207 R i, Oe 


18 "21 . 240 22m 
— = = = 6, &,, 
gr hge Vere oee ‘ 
and any two corresponding values of «, y, being substituted in the 
given equation, 

5a — 3y = 12 


will satisfy it equally well: hence, there are an infinite number 
of values for « and y which will satisfy the equation, and conse- 
quently, the problem is indeterminate ; that is, it admits of an in- 
finite number of solutions. 

If, however, we impose a second condition, as for example, 
that the sum of the two quantities shall be equal to 4, we shall 
have a second equation, 

e-+- y= 4; 
and this, combined with the equation already considered, will 
give determinate values for x and y. 

If we have two equations, involving three unknown quantities, 
we can eliminate one of the unknown quantities, and thus ob- 
tain an equation containing two unknown quantities. This equa: 
tion, like the preceding, would be satisfied by an imfmite num: 
ber of values, attributed in succession, to the unknown quanfi- 
ties. Since each equation expresses one condition of a problem 
therefore, in order that a problem may be determinate, its enum: 
ciation must contain at least as many different conditions as ther 
are unknown quantities, and these conditions must be such, that eac} 
of them may be expressed by an independent equation; that 1s, at 
equation not produced by any combination of the others of the system 

If, on the contrary, the number of independent equations ex 
ceeds the number of unknown quantities involved in them, | 
conditions which they express cannot be fulfilled. 


“ 
OHAP. IV.) EQUATIONS OF THE FIRST DEGREE. 85 
i 
For example, let it be required to find two numbers such that 


heir sum shall be 100, their difference 80, and their product 700. 
The equations expressing these conditions are, 


2+ y= 100 
xe—-y= 80 
| and esi S700. 


' Now, the first two equations determine the values of x and y, 
nz., 
2-90 aid y—10. 


‘The product of the two numbers is therefore known, and equal 
o 900. Hence, the third condition cannot be fulfilled. 

Had the’ product been placed equal to 900, all the conditions 
‘vould have been satisfied, in which case, however, the third would 
‘hot have been an independent equation, since the condition ex- 
pressed by it, is implied in the other two. 








EXAMPLES. 

| 
1. Given 2r-+3y=16, and 3x—2y=11 to find the values 

f x and y. ANS: © ==. 5, ei 


£ OR nay 9 Sm 9 2Y 61 
2. Given ; - A 50" and if -- - 150 ) e 
1 1 
values of w and y. Ans. x = a en 3" 


3. Given + + 7y= 99, and : + 7x = 51 to find the values 
if x and y. Ans.i% = 1,) y= dA 





| _2 
) 4. Given S—w=54+8, and 44 i = 427 
I) find the values of # and y. — Ans. « = 60, y = 40. 


e+ yt 2=29 


ea oe to find a, y, and z~ 


1 Pl 
r= “a —z=—10 
oe Tl Ie ng 


ATs, = 0, Y= 0, Zine 


86 ELEMENTS OF ALGEBRA. [CHAP. IV, 


22 4 4y — 32 = 22 
4xn— 2y+ 52=18 to find a, y, and 2 
6e+ Ty—- 2=63 

Ang, 2= 3, Yea, 13 = & 


6. Given 


7. Given ZV to find a, y, and z. 


Ans. 2 = 12; =a ae = 30. 


("Te — 22 su V7 
Ayn Oe te dd 


8. Given 5y— 34— 2u= 8 to find a, y, Z, 4, 
4y— 3u+ 2= 9 and ¢. 
32 ++ ((8u == 33 


Ans. = 2, y=4, 2= 3, Boog, b= he 


QUESTIONS. 


1. What fraction is that, to the numerator of which, if 1 be 
added, its value will be one third, but if one be added to its de- 
nominator, its value will be one fourth. 


Let the fraction be represented by a 


Then, by the question art = = and aa = 2 
. Whence 3r+3=y, and 4a=y+1. 

Therefore, by subtracting, #w2—3=1 or «= 4; | 

and 3X 4$-3= 15>. > 


2. A market woman bought a certain number of eggs at 2 for 
a penny, and as many more, at 3 for a penny, and having sold 
them again altogether, at the rate of 5 for 2d., found that she 
had lost 4d.: how many éggs had she? f 


Dd” : i 


; 


hy 
‘ 


CHAP. IV.] EQUATIONS OF THE FIRST DEGREE. 87 


Let 22 = the whole number of eggs; 
then «= the number of eggs of each sort; 
I 
and rag the cost of the first sort ; 
1 
and fete the cost of the second sort; 
| 4 
But Bir 2 22 Qe toms 


| 4 
‘hence, “ the amount for which the eggs were sold. 


Hence, by the question, 


1 1 4a 
: 
therefore 152 + 102 — 24” = 120. 
Or, x == 120 the number of eggs of 


each sort. 


3. A person possessed a capital of 30,000 dollars, for which he 
‘drew a certain interest per annum; but he owed the sum of 
20,000 dollars, for which he paid a certain interest. ‘The inter- 
est that he received exceeded that which he paid by 800 dollars. 
Another person possessed+ $35,000, for which he received interest 
at the second of the above rates; but he owed 24,000 dollars; 
for which he paid interest at the first of the above rates. The 
‘interest that he received exceeded that which he paid by 310 
‘dollars. Required the two rates of interest. 

Let « and y denote the two rates of interest: that is, the in- 
terest of $100 for one year. 

To obtain the interest of $30,000 at the first rate, denoted by a, 
we form the proportion 






30,000x 
De cs .: 30 ,000 apie Lae ie Md or 3002. 
And for the interest $20,000, the rate being y, 
20,000 
100 : y :: 20,000 : : cares or 200y. 


But from the enunciation, the difference between these two in 
|terests is equal to 800 dollars. 
We have, then, for the first equation of the problem, 


300x — 2007 = 800. 


88 ELEMENTS OF ALGEBRA. [CHAP. IV. 


By expressing the second condition of the problem algebraically, 
we obtain the other equation, 


350y — 2402 = 310. 


Both members of the first equation being divisible by 100, and 
those of the second by 10, we may put the following, in place 
of them: 


32 — 2y = 8, .35y — 24% = 31. 


To eliminate a, multiply the first equation by 8, and then add 
it to the second; there results 


- 


19y = 95,. whence ¥ = 3, 


Substituting for y its value in the first equation, this equation 
becomes 


32 —10=8, whence x= 6; 


Therefore, the first rate is 6 per cent., and the second 5. 


Verification. 


$30,000, placed at 6 per cent., gives 300 X6 = $1800. 
~ $20,000 do. 9) do. 200 x 5 = $1000. 
And we have 1800 — 1000 = 800. 


The second condition can be verified in the same manner. 


4. There are three ingots formed by mixing together threg 
metals in different proportions. 

One pound of the first contains 7 ounces of, silver, 3 ounces of 
copper, and 6 ounces of pewter. 

One pound of the second contains 12 ounces of silver, 3’ ounces 
of copper, and 1 ounce of pewter. 

One pound of the third contains 4 ounces of silver, 7 ounces 
of copper, and 5 ounces of ‘pewter. 

It is required to form from these three, 1 pound of a fourth 
ingot which shall contain 8 ounces of silver, 32 ounces of copper, 
and 44 ounces of pewter. 

Let «= the number of ounces taken from the first. . 

y = the number of ounces taken from the second. 


z= the number of ounces taken from the third. 


CHAP. IV.] EQUATIONS OF THE FIRST DEGREE. 89 


, Now, since 1 pound or 16 ounces of the first ingot contains 7 


q f : ; 1 

ounces of silver, one ounce will contain 16 of 7 ounces: that 

ils, i6 ounces ; and 
; : it ; 

x ounces will contain 18 ounces of silver. 

; ru kee ' 

y ounces will contain te ounces of silver. 

]s 
i } 4z : 

z ounces will contain ié ounces of silver. 


But since 1 pound of the new ingot is to contain 8 ounces of 
isilver, we have 
| 12 reais va ote 8: 
or, reducing to entire terms, 

Te + 12y + 42 = 128. 

For the copper, 32 + 38y+72= 60; 
and for the pewter, 62+ y+5z= 68. 

As the co-efficients of y in these three equations, are the most 
simple, we will eliminate this unknown quantity first. 

Multiplying the second equation by 4 and subtracting me first, 


| ie 
, 5a + 242 = 112. 


Multiplying the third equation by 3 and subtracting the second, 
gives 
1l5a-+ 82 = 144. 
Multiplying the last equation by 3 and subtracting the first, gives 
402 = 320, 
whence f= 8 
| Substituting this value of » in the equation 
je ° 5x + 242 = 112, 
“tt becomes 40 + 242 = 112, whence z=3. 
Lastly, the two values «=8 and z= 3, being substituted 


in the equation 
62 + y +,5z = 68 


, give 48+y+15=68, whence y=5 





90 ELEMENTS OF ALGEBRA [CHAP. IV. 


Therefore, in order to form a pound of the fourth ingot, we 
must take 8 ounces of the first, 5 ounces of the second, and 3 
_of the third. 


Verification. 


If there be 7 ounces of silver in 16 ounces of the first ingot, 
in 8 ounces of it, there should be a number of ounces of silver 
x8 


expressed by i6 





Aj eae 4 KaeBy 

- and “Se will, express the quan- 
tity of silver contained in 5 ounces of the second ingot, and 3 
ounces of the third. Now, we have 


TeCB. (38 o6 5 eee 128 
a i 
16 a 16 a 16 16 . 


therefore, a pound of the fourth ingot contains 8 ounces of silver, 
as required by the enunciation. The same conditions may be verl- 
fied relative to the copper and pewter. 


In like manner, 





5. What two numbers are those, whose difference is 7, and 
sum 33? Ans. 13 and 20. 
6. To divide the number 75 into two such parts, that three 
times the greater may exceed seven times the less by 15. 
: Ans. 54 and 21. 
7. In a mixture of wine and cider, } of the whole plus 25 gal- 
lons was wine, and } part minus 5 gallons was cider; how many 
gallons were there of each ? 
Ans. 85 of wine, and 35 of cider. 
8. A bill of £120 was paid in guineas and moidores, and the 
number of pieces of both sorts that were used was just 100; if 
the guinea were estimated at 2ls., and the moidore at 27s., how 
many were there of each? Ans. 50 of each. 


9. Two travellers set out at the same. time from London and 
York, whose distance apart is 150 miles; one of them goes 8 
miles a day, and the other 7; in what time will they meet? 

Ans. In 10 days. | 

10. At a certain election, 375 persons voted for two candi- 
dates, and the candidate chosen had a majority of 91; how many 
voted for each? Ans. 233 for one, and 142 for the other. 


é | 
. { 


st. os. 


“4 at Se 








(ar. 

i ; & 
Ee : 

ms 

bi 


- CHAP. IV.] EQUATIONS OF THE FIRST DEGREE. 9] 


11. A’s age is double of B’s, and B’s is triple of C’s, and the 


‘sum of all their ages is 140; what is the age of each? 


Ans. A’s = 84, B’s= 42, and C’s = 14. 
12 A person bought a chaise, horse, and harness, for £60; 
the horse came to twice the price of the harness, and the chaise 
to twice the price of the horse and harness; what did he give 


for each? £13 6s. 8d. for the horse. 
Ans £ 6 13s. 4d. for the harness. 
£40 for the chaise. 


13. Two persons, A and B,,have both the same income. A 
saves 1 of his yearly; but B, by spending £50 per annum more 
than A, at the end of 4 years finds himself £100 in debt; what 
is the income of each? Ans. £125. 


14. A person has two horses, and a saddle worth £50; now, 


if the saddle be put on the back of the first horse, it will make 


his value double that of the second; but if it be put on the back 
of the second, it will make his value triple that of the first; 


what is the value of each horse ? 


Ans. One £30, and the other £40. 
15. To divide the number 36 into three such parts, that 4 of 
the first, } ofethe second, and 1 of the third, may be all eet to 
each hk Ans. 8; 12;) and 16, 
16. A footman agreed to serve his master for £8 a year and 
a livery, but was turned away at the end of 7 months, and re- 


i ceived only £2 13s. 4d. and his livery; what was its value ? 


; Ans. £4 16s. 
17. To divide the number 90 into four such parts, that if the 
first be increased by 2, the second diminished by 2, the third 


) multiplied by 2, and the fourth divided by 2, the sum, difference, 


-prod.ict, and quotient so obtained, will be all equal to each other. 
Ans. The parts are 18, 22, 10, and 40. 
18. The hour and minute hands of a clock are exactly together 
at 12 o’clock; when are they next together? 
Ans. 1h. 53% min. 


19. A man and his wife usually drank out a cask of beer in 


‘m2 days ; but when the man was from home, it lasted the woman 


30 days ; how many days would the man be in drinking 1 it alone ? 
Ans. 20 ‘days. 


32 , 8 MBLEMENTS OF ALGEBRA. [CHAP. TV. 


20, If A and B together can perform a piece of work in 8 
_days,; A and C together in 9 days, and B and C in 10 days; 
how many days would it take each person to perform the same 
work: alone ? Ans. A 1434 days, B 1728, and C 2324. 
21 A laborer can do a certain work expressed by a, in a time 
expressed by 5; a second laborer, the work ¢ in a time d; a 


third, the- work e in a time f. Required the time it would take 


the three laborers, working together, to perform the work g. 


bdfg 
UG sich aon 
Gores) b= 4) e = 35; d= 6 |e = Aes f=i2|g=191; 
_ w@ will be found equal to 12. 

22. 1f 32 pounds of sea water contain 1 pound of salt, how 
much fresh water must be added to these 32 pounds, in order 
that the quantity of salt contained in 32 pounds of the new mix- 
ture shall be reduced to 2 ounces, or 3 of a pound? 

Ans. 224 lbs. 

23. A number is expressed by three figures; the sum of these 
figures is 11; the figure in the place of units is double that in 
the place of thunivbae’ and when 297 is added to this number, 
the sum obtained is expressed by the figures of this number re- 
versed. What is the number? Ans. 326. 


24. A person who possessed 100,000 dollars, placed the greater 


part of it out at 5 per cent. interest, and the other part at 4 per 


cent. The interest which he received for the whole pperes 


to 4640 dollars. Required the two parts. 
Ans. $64,000 and $36, 000. 
25. A person possessed a certain capital, which he placed out 


t 


at a certain interest. Another person possessed 10 000 dollars | 
more than the first, and putting out his capital 1 per cent. more 
advantageously, had an income greater by 800 dollars. A. third, 
possessed 15,000 dollars more than the first, and putting out his) 
capital 2 per cent. more advantageously, had an income greater 


by 1500 dollars. Required the capitals, and the three rates of 


inter 
- Sums at interest, $30,000, $40,000, $45, 000. 


A 


q 
Rates of interest, ie (ee 6 per cent “4 
is | 





CHAP. IV.] EQUATIONS OF THE FIRST DEGREE. 93 


_ 26. A banker has two kinds of money; it takes a pieces of 
‘the first to make a crown, and } of the second to make the 
same sum. Some one offers him a crown for c- pieces. How 
many of each kind must the banker give him? 
Ans. 1st kind, aie a 2d kind, ad iw Neg 


; 








27. Find what each of three persons, A, B, C, is worth, know- 
‘ing, lst, that what A is worth added to 7 times what B and C 
are worth, is equal to p; 2d, that what B is worth added to m 
‘times what A and C are worth,is equal to g; 3d, that what C is 
worth added to m times what A and B are worth, is equal to r. 
If we denote by # what A, B, and C, are worth, we introduce 
jinto the calculus an auxiliary unknown quantity, and resolve the 
‘question in a very simple manner. The term calculus, in its gen- 
‘eral sense, denotes any operation performed on algebraic quantities. 

28. Find the values of the estates of six persons, A, B, C, D, 
E, F, from the following conditions: 1st. The sum of the estates 
of A aud B is equal to a; that of C and D is equal to 6; and 
‘that of E and F is equal to c. 2d. The estate of A is worth m 
times that of C; the estate of D is worth n times that of E, and 
‘the estate of F is worth p times that of B. 

This problem may be resolved by means of a single equation, 
probing but one unknown quantity. 

Explanation of Negative Results. 





104, The algebraic signs are an abbreviated language. They 
Indicate certain’ operations which are to be performed on the quan- 
tities before which they are placed. ‘ 

The operation indicated by a particular sign, must be per- 
formed on every quantity before which the sign is placed. In- 
'deed, the principles of Algebra are all established upon the 
supposition, that each particular sign which is employed means 
always the same thing; and that whatever it requires is strictly 
performed. ‘Thus, if the sign of a quantity is +-, we understand 
that the quantity is to be added; if the sign is —, we under-_ 
istand that it is to be subtracted. , 

For example, if we have — 4, it indicates that this 4 is to 
he subtracted from some other number, or that it is the result ae 
a Buntraqaon but partially made. 





94 ELEMENTS OF ALGEBRA. [CHAP. IV. 


If it were required to subtract 20 from 16, the subtraction could 
not be made by the rules of arithmetic, since 20 is greater than 16. 
By observing that 


a0 16 a 
we may express the subtraction thus, 
16 —20 = 16—16—4 = —4. 


We thus make the subtraction of 20 from 16 as far as it is 
possible, and obtain a remainder 4 with a minus sign, which in- 
dicates that 4 is still to be treated as a subtractive quantity. 

To show the necessity of giving to this remainder its proper 
sign, let us suppose that 10 is to be added to the difference of 
16 — 20; or what is the same thing, that 20 is to be subtracted 
from 26. 


The numbers would then be written 
16. 20. yee 4 


+ 10 — +10 
26 —20=-+ 6; 


and had the — sign not been preserved in the first subtraction, 
the second result would have been + 14 instead of +6. 


105. If the sum of the negative quantities in the first member 
of the equation, exceeds the sum of the positive quantities, the 
second member of the equation will be negative, and the verifica- 
tion of the equation will show it to be so. 


For example, if a—b=c, 
and we make a=15 and 6=—18, c will be = —3. 


Now, the essential sign of ¢ is different from its algebraic sign 
in the equation. This arises from the circumstance, that the 
equation 

a—b=ce 
expresses generally, the difference between a and 6, without in- 
dicating which of them is the greater. When, therefore, we at- 
tribute particular values to @ and 8, the sign of ¢, as well as its 
value, becomes known. 

We will illustrate these remarks by a few examples. 














HAP. IV.] EQUATIONS OF THE FIRST DEGREE. 95 


. 1. To find a number, which added to the number 38, will give 
sum equal to the number a. 


Let «= the required number. 
Then, by the conditions 


z+b=a, whence rx =a—b. 


This expression, or formula, will give the algebraic value of 
in all the particular cases of this problem. 


For example, let @—=47 and 6—29; 


then, v= 47 — 29 = 18. 
Again, let @o—24 \and 6 = 3); 
then, x= 24—31= — 7. 


This last: value of x, is called a negative solution. How is it 
|) be interpreted ? 

If we consider it as a purely arithmetical result, that is, as 
\Yising from a series of operations in which all the quantities are 
‘garded as positive, and in which the terms add and subtract 
inply, respectively, augmentation and diminution, the problem will 
bviously be impossible for the last values attributed to a and ); 
or, the number #4 is already greater than 24. 

| Considered, however, algebraically, it is not so; for we have 
jyund the value of x to be —7, and this number added, in the 
lgebraic sense, to 31, gives 24 for the algebraic sum, and there- 
re satisfies both the equation and enunciation. 


| 2. A father has lived a number a of years, his son a number 
|f years expressed by b. Find in how many years the age of 
jie son will be one fourth the age of the father. 

reet  ’ x” = the required number of years. 

Then a+a= the age of the | at the end of the re- 
| and b+a= the age of the son quired time. 
Hence, by the question 


ore ab+e; whence, poo. 








Suppose a—54, and b=9; then « = TS ba 















96 ELEMENTS OF ALGEBRA. (CHAP. IV. 


The father being 54 years old, and the son 9, in 6 years the 
father will be 60 years old, and his son 15; now 15 is the 
fourth of 60; hence, «= 6 satisfies the enunciation. 


Let us now suppose a@=45, and 6=15; 








45 — 60 (i 
then, x= ——— = — BD. 
3 
If we substitute this value of « in the equation of condition, 
a+a 
iv ta b+ 2, 
, 45 — 5 
we obtain, ae 15 — 5; 
and (OZ 
Hence, — 5 substituted for «x, verifies the equation, and therefor« 


is a true answer. 

Now, the positive result which was obtained, shows that th 
age of the father will be four times that of the son at the ex 
piration of 6 years from the time when their ages were cop 
sidered; while the negative result, indicates that the age of th 
father was four times that of his son, 5 years previous to th 
time when their ages were compared. 

The question, taken in its general, or algebraic sense, demand: 
the time, at which the age of the father was four times that o 
the son. In stating it, we supposed that the age of the fathe 
was to be augmented; and so it was, by the first supposition. Bu 
the conditions imposed by the second supposition, required thi 
age of the father to be diminished, and the algebraic result con 
formed to this condition, by appearing with a negative sign. I 
we wished the result, under the second supposition, to have : 
positive sign, we might alter the enunciation by demanding, ho 
many years since the age of the father was four times that of th 
son. 


If « — the number of years, we shall have 
= b—2: hence, “2 = i 


If a=45 and }=15, «x will be equal to 5. 








Reasoning from analogy, we establish the following genera 
principles, 


/ CHAP. Iv.] EQUATIONS OF THE FIRST DEGRER. 97 


Ist. Every negative value found for the unknown quantity in a 
problem of the first degree, will, when taken with its proper sign, 
verify the equation from which it was derived. 





2d. That this negative value, taken with its proper sign, will alse 
satisfy the enunciation of the problem, understood in its algebrate 
sense. 

3d. The negative result shows that the enunciation is wnpossible, 
regarded in tts arithmetical sense. The language of Algebra de- 
‘tects the error of the arithmetical enunciation, and indicates the gen- 
eral relation of the quantities. 
Ath. The negative result, considered without reference to its Sign, 
may be regarded as the answer to a problem of which the enuncia- 





ten only differs from the one proposed in this: that certain quan- 
‘ties which were additive have become subtractive, and reciprocally. 


106. As a further illustration of the “change which an alge- 
onic sign may produce in the enunciation of a problem, let us 
fesume that of the laborer (page 76). 

, Under the supposition that the laborer receives a sum c, we 
ave the equations 
ime + ree: whence, eee aie Heed. 

ax — by =c a+b a+b 
i If at the end of the time, the laborer, instead of receiving a 
sum c, owed for his board a sum equal to c, then, by would be 
‘eater than ax, and under this supposition, we should have the 
equations | 
{ e+y=n and ax—by= —c. 

_ Now, it is plain that we can obtain immediately the values of 
sand y, in the last equations, by merely changing the sign of 
_ In each of the values found from the equations above; this 
‘ives 





bn —c an+ec 
he 1b: al ath 
| The results for both enunciations, may be compreiiended in the 
ame formulas, by writing 





bn+e ; an=ec 

Pa pe PER 
_ The double sign +,.. read plus or minus, and =, is read, mi 
us.or plus. The upper signs vurrespond to the case in which 


a ae Se 


98 ELEMENTS OF ALGEBRA. [CHAP. IV. 


the laborer received, and the lower signs, to the case in which 
he owed a sum c. These formulas also comprehend the case in 
which, in a settlement between the laborer and his employer, 
their accounts balance. This supposes c= 0, which gives 

bn an 


2= oS 


Discussion of Problems. Explanation of the terms 
Nothing and Infinity. 


107. When a problem has been resolved generally, that is, by 
means of letters and signs, it is often required to determine what 
the values of the unknown quantities become, when particular sup- 
positions are made upon the quantities which are given. The 
determination of these values, and the interpretation of the pe- 


culiar results obtained, form what is called the discussion of the 
problem. 


The discussion of the following question presents nearly all 
the circumstances which are met with in problems of the first 
degree. 


108. Two couriers are travelling along the same right line and 
in the same direction from R’ toward R. The number of miles 
travelled by one of them per hour is expressed by m, and the 
number of miles travelled by the other per hour, is expressed 
by n. Now, at a given time, say 12 o'clock, the distance be- 
tween them is equal to a number of miles expressed by a: re- 
quired the time when they will be together. 


jane A B R. 





At 12 o’clock suppose the forward courier to be at B, the other 
at A, and R to be the point at which they will be together. 
Then, AB =a, their distance upart at 12 o’clock. 


Let t= the number of hours which must elapse, be- 
fore they come together ; 
and a= the distance BR, which 1s to be passed over 


by the forward courier. 
Then, since the rate per hour, multiplied by the number of 
hours, will give the distance passed over by each, we have, 
tx m=a=atas AR 
EX nm 2x = BH 


f 
li 


; 


‘HAP. 1v.] EQUATIONS UF THE FIRST DEGREE. 99 


. 


| Hence by subtracting, 


t(m—n) =a, 


. 


and hence, see ee 
m—n 

_ Now, so long as m>n, ¢ will be positive, and the problem 
vill be solved in the arithmetical sense of the enunciation. For, 
f m>n, the courier from A will travel faster than the courier 
‘om B, and will therefore be continually gaining on him: the 
terval which separates them will diminish more and more, un- 
‘1 it becomes 0, and then the couriers will be found upon the 
‘ame point of the line. 

' In this case, the time ¢, which elapses, must be added to 12 
clock, to obtain the time when they are together. 

! But, if we suppose m< _n, then, m—~n will be negative, and 
‘ae value of ¢ will be negative. How is this result to be inter- 
reted ? 

It is easily explained from the nature of the question, which 
‘onsidered in its most general sense, demands the time when the 


















yuriers are together. 

Now, under the second supposition, the courier which is in ad- 
‘ance, travels the fastest, and therefore will continue to separate 
‘imself from the other courier. At 12 o’clock the distance be- 
‘veen them was equal to a: after 12 o’clock it is greater than a; 
fad as the rate of travel has not been changed, it follows that 
Srevious to 12 o’clock the distance must have been less than a. 
/t a certain hour, therefore, before 12, the distance between them 
\ust have been equal to nothing, or the couriers were together at 
lyme point R’. The precise hour is found by subtracting the 
‘ilue of ¢ from 12 o'clock. 
| This example, therefore, conforms to the general principle, that, 
| the conditions of a problem are such as to render the unknown 
lvantity essentially negative, it will appear in the result with the 
linus sign, whenever it has heen regarded as positive in the enun- 
| ation. 
| If we wish to find the distances AR and BR, passed over by 
/e two couriers before coming together, we may take the equation 
a 


aa 


m—-7n 


100 ELEMENTS OF ALGEBRA. [CHAP. IV 


and multiply both members by the rates of travel respectively : 
this will give 





ma 
AR =m = and 
m—n 
na 
BR = nt = ———. 
m—n 
ma 
Also, AR’ = — mt = ——— 
| m—n 
na 
and BR’ = — at = ———;; 
m—-n 


from which we see, that the two distances AR and BR, will 
both be positive when estimated toward the right, and that AR’ 
and BR’ will both be negative when estimated im the contrary 
direction. 

109. To explain the terms nothing and infinity, let us consider 


the equation 


a 
|, saa 





m—n 

If the couriers travel at different rates, m—n will be a finite 
quantity, and its sign will depend on the relative values of m 
and n. Designate this quantity by A. 

Now, if we suppose a@=0, we shall have 

t= or th <A =e 
an equation which can only be satisfied by making t = 0. 

To interpret this result, let us go back to the enunciation of 
the problem. If a=0, the couriers are together at 12 o’clock; 
and since they travel at different rates, they can never be again 
together: hence, ¢ can have no other value than 0. Therefore, 
we conclude that, the quotient of 0 divided by a fate quantity, is 0. 

110. Let us resume the’ equation 

oi Dehra ; 
m—n 
If in this equation we make m=n, then m—n=0O, and 
the value of ¢ will reduce to | 
a 
=a 0 an:? x 0 =a 
an equation which cannot be satisfied for any finite value of t 


| 
| CHAP. IV.] _ NOTHING AND INFINITY. 101 


. In order to interpret this new result, let us go back to the 
‘enunciation of the question. We see at once, that it is abso- 
\lutely impossible to satisfy the enunciation for any finite value 
for ¢; for, whatever time we allow to the two couriers, they can 
‘Tever come together, since being once separated by an _ inter- 
val a, and travelling equally fast, this interval will always be pre- 
served. 


a 
| Hence, the result, > may be regarded as a sign of impossi- 


bility for any finite value of ¢. 
_ Nevertheless, algebraists consider the result, 


t —— 0” 
as forming a species of value, to which they have given the 
name of infinite value, for this reason: 

When the difference m—n, without being absolutely nothing, 
is supposed to be very small, the result 








a 
t= ————.. 
{ m—n 
is very great. 

Take, for example, m—n= 0,01. 

a a 
h = OS ss ————- = ay 
Then t as 0101 100a 
Again, take m—n=0.001, and we have 
a a 
———_—— = —— = 1000a. 
: m—n 0.001 ‘ 


In short, if the difference between the rates is not zero, the 
couriers will come together at some pomt of the line, and the 


ished. 
| Hence, ‘from analogy, if-the difference between the rates is less 


jthan any assignable number, the time expressed by 
a a 
t= ——_ = 


i) 
m—n 0 


\will be greater than any assignable or finite number. Therefore 
for brevity, we say, when m— n=O, the result, 


102 ELEMENTS OF ALGEBRA. (CHAP. IV. | 


becomes equal to infinity, which we designate by the charac- | 


ter o. | 
Hence we conclude, that a finite quantity divided by 0, gives | 
a quotient greater than any assignable quantity, which we call, | 













INFINITY. 


111. Again, let A represent any finite number: then, since thie | 
value of a fraction increases as its numerator becomes greater| 
with reference to its denominator, the expression 


A 
0’ 
is a proper symbol to represent an infinite quantity; that is, aj 
quantity greater than any assignable quantity. | 
Since the value of a fraction diminishes as its denominator be- | 
comes greater with reference to its numerator, the expression 
A 
re 
is a proper symbol for a quantity less than any assignable quan: | 


tity. Hence, 


0 and o 


are synonymous symbols ; and so likewise, are 


cis and 0. 
or) 


We have been thus particular in explaining these ideas of mn-| 
finity, because there are some questions of such a nature, that| 
infinity may be considered as the true answer to the enunciation.’ 

In the case just considered, where m= 2, it will be perceived 
that there is not, properly speaking, any solution in finite and de- 
terminate numbers ; but the value of the unknown quantity is found) 
to be infinite. 


112. If, in addition to the hypothesis m = n, we also suppose 
0 ‘ 
that a = 0, we have ¢ = oo tx 0=0; a result which will 


be satistied by any value of ¢. 

To interpret this result, let us consider again the enunciation. 
from which it is perceived, that if the two couriers travel equally 
fast, and are once at the same point, they ought, ever after, to 


be together, and consequently the required time is entirely unde- 
. , 


\ OHAP. IV.] NOTHING AND INFINITY 103 






ao. Osha ; 
‘termined. Therefore, the expression > 3%.m this case, the 


ssymbol of an indeterminate quantity. 

' The preceding suppositions are the only ones that lead to re- 
‘markable results; and they are sufficient to show to beginners 
the manner in which the results of Algebra answer to all the cir- 
‘cumstances of the enunciation of a problem. 


113. It should be observed, that the expression im is not a 


certain symbol of indetermination, but frequently arises from the 
existence of a common factor in each term of the fraction, which 
-factor becomes nothing, in consequence of a particular hypothesis. 
_ For example, suppose the value of the unknown quantity to be 
a’ — 53 
= OnE Pe je 
If, in this formula, @ is made equal to 0d, there results 
0 
as 


0 
_ But observe (Art. 48), that 
| a3 — b3 — (a — b) (a2 + ab + B) 

and a® — hb? — (a — b) (a+ J), 
hence, we have 
: a (a + cb + BY 
ee a+ J) 

_ Now, if we suppress the common factor a — b, and then sup- 
‘pose a= b, we shall have ; 
Od Gla sk (0%... 3a? 3a 

















atb 7 eae 
Let us suppose, that in another example, we have 
ese 
a (a —b) 
| If we suppose a = b, we have 
r 0 
ie 
0 


If, however, we suppress the factor common to the numerator 
and denominator, in the value of x, we have, 
(a+b)(a—b) atb_ 2b “ 
(4 | ioe SS es 
(a—b)(a—b) .a—b 0 











104 ELEMENTS OF ALGEBRA. [CHAP. IV 


Therefore we conclude, that before pronouncing upon the true 
value of the fraction, 
0 
0 
it Is necessary to ascertain whether the two terms do not contain 
a common factor. If they do not, we conclude that the fraction is 
really zndeterminate. If they do contain one, suppress it, and then 
make the particular hypothesis; this will give the true value of 


the fraction, which will assume one of the three forms 
A ANG 
3? "OF Tike 
that is, it will be determinate, infinite, or indeterminate. 
This observation is very useful in the discussion of problems. 


Of Inequalities. 


114. In the discussion of problems, we have often occasion to 
suppose quantities unequal, and to perform transformations upon 
them, analogous to those executed upon equalities. We some- 
times do this, to establish the necessary relations between the 
given quantities, in order that the problem may be susceptible of 
a direct, or at least, of a real solution. We often do it, to fix the 
limits between which the particular values of certain given quan- 
tities must be found, in order that the enunciation may fulfil a 
particular condition. Now, although the principles established for 
equations are, in general, applicable to inequalities, there are nev- 
ertheless some exceptions, of which it is necessary to speak, in 
order to put the beginner upon his guard against some errors that 
he might commit, in making use of the sign of inequality. ‘These 
exceptions arise from the introduction of negative expressions into 
the calculus, as quantities. 

In order to be clearly understood, we will give examples of | 
the different transformations to which inequalities may be su 
jected, taking care to point out the exceptions to which onal 
transformations are liable. ¢ 


J : 
115. Two inequalities are said to subsist in the same sense, 
when the greater quantity stands at the left in both, or at the 
right in both; and in a contrary sense, when the greater quan- 


tity stands at the right in one, and at the left in the other. 


« 


\ : 
9 


‘UHAP. IV.]} UF INEQUALITIES. 105 


peed hus, © 25 >20° and 18> 10, or 6<(8 and 7< 9, 


are inequalities which subsist in the same sense; and the in- 
‘equalities 
15>13 and 12< 14, 
: Wi vis 
subsist In a contrary sense. 
( 


{ aed. If we add the same quantity to both members of an inequality, 
or subtract the same quantity from both members, the resulting tn- 
equality will subsist in the same sense. 


' Thus, take 8 >6; by adding 5, we still have 


. 


8+5>6+5; 
and subtracting 5, we have 
i 

8—5>6—5. 





When the two members of an inequality are both negative, that 
one is the least, algebraically considered, which contains the great- 
est number of units. Thus, — 25 << — 20; and if 30 be added 
to both members, we have 5< 10. This must be understood 
‘entirely in an algebraic sense, and arises from the convention be- 
fore established, to consider all quantities preceded by the minus 
sign, as subtractive. | 
.' The principle first enunciated, serves to transpose certain terms 
from one member of the inequality to the other. Take, for ex- 
ample, the inequality 


a2 + hb? > 3b? — 2a?; 
‘there will result, by transposing, 


a® + 2a? > 362 — 6, or 3a? > 262. 






| 2. If two inequalities subsist in the sume sense, and we add them 
member to member, the resulting inequality will also subsist in the 
same sense. 

+ Thus, add ¥v, Le Picea dye >]; and 
vhere results a +e “a e>b+id+f. 

| But this is Wer always a case, when we subtract, member from 
member, ‘two inequalities established in the same sense. 

Let there be the two inequalities 4<7 and 2 < 3, we have 


4—2 or 2< 7—3 or 4. 


106 | ELEMENTS OF ALGEBRA. [CHAP. IV. 


But if we have the inequalities 9< 10 and 6 < 8, by sub- 


tracting we have 
9—6 or 3>10—8 or 2. 

We should then avoid this transformation as much as possible, 
or if we employ it, determine in which sense the resulting in- 
equality exists. 

3. If the two members of an inequality be multiplied by a positive 
number, the resulting inequality will exist in the same sense. 

Thus, a<b, will give 3a < 30; 

and, —a<b, —3¢ << =e 

This principle serves to make the denominators disappear. 
Aes 2. Wee 


2d Mg 3a 





From the inequality , we deduce, by mul- 


tiplying by 6ad, 
3a (a? — b?) > 2d (c? — d?), 
and the same principle is true for division. 


But, when the two members of an inequality are multiphed a 
divided by a negative number, the inequality will subsist in a con 
trary sense. 

Take, for example, 8 > 7; multiplying by — 3, we have 


—24< —2l. 





8 
3°, Cem 
Therefore, when the two members of an inequality are multi- 
plied or divided by a number expressed algebraically, it is ne- 
cessary to ascertain whether the mu/tiplier or divisor is negative; 
for, in that case, the inequality will exist in a contrary sense. 


In like, manner, 8 >7 gives 


4. It is not permitted to change the signs of the two members of. 
an inequality, unless we establish the resulting inequality in a con- 
trary sense; for this transformation is evidently the same as mul- 
tiplying the two members by — 1. 





5. Both members of an inequality between positive numbers can 
be squared, and the inequality will exist in the same sense. | 
Thus, from 5>3, we deduce, 25 >9; from a+6> 0, we 
find | 
(a + 5)? > e?. | 


CHAP. IV.] OF INEQUALITIES. 107 


6. When the signs of both members of, the inequality are not 
known, we cannot tell before the operation is performed, in which 
‘ sense the resulting inequality will exist. 

; For example, —2<3 gives (—2)? or 4<9; but 3>—5 
; gives, on the contrary, (3)? or 9< (— 5)? or 25. 

| We must, then, before squaring, ascertain the signs of the two 
. members. | 


| EXAMPLES. 
| 1. Find the limit of the value of « in the expression 
5a — 6 > 19. ARS, Qian. 


2. Find the limit of the value of « in the expression 





4 
32 + <2 — 30> 10 Ans. x > 4. 


3. Find the limit of the value of « in the expression 





1 1 1 grata as 
et-zete te > Ans. « > 6. 
4. Find the limit of the value of 2 in the inequalities 
2 
~ + br — ab> = 
bx b? 
ae ax +ab< ign 


5. The double of a number diminished by 5 is greater than 
‘25, and triple the number diminished by 7, is less than double 
‘the number increased by 13. Required a number which shall 
‘Satisfy the conditions. 
| By the question, we have 


Qn — 5 > 25. 
32 — 7 < 2x + 13. 







Resolving these inequalities, we have #>15 and 2x < 20. 
Any number, therefore, either entire or fractional, comprised be- 
| tween 15 and 20, will satisfy the conditions. 


J08 ELEMENTS OF ALGEBRA. [CHAF. V 


CHAPTER V. 


EXTRACTION OF THE SQUARE ROOT OF NUMBERS.—FORMATION OF 
THE SQUARE AND EXTRACTION OF THE SQUARE ROOT OF ALGE- 
BRAIC QUANTITIES.—CALCULUS OF RADICALS OF THE SECOND 
DEGREE. f 


116. THE square or second power of a number, is the product 
which arises from multiplying that number by itself once: for 
example, 49 is the square of 7, and 144 is the square of 12. 

The square root of a number, is that number which multiplied 
by itself once will produce the given number. Thus, 7 is the 
Square root of 49, and 12 the square root of 144.5 [On tod —— 4 
and 12 x'12 = 144. 

The square of a number, either entire or fractional, is easily 
found, being always obtained by multiplying the number by itself 
once. The extraction of the square root is, however, attended 
with some difficulty, and requires particular explanation. 

The first ten numbers are, 


1, 25.) 85 4s Os 6 97, 
and their squares, 

1, 4, 9, 16, 25, 36, 49, 647eRipeenone 
and reciprocally, the numbers of the first line are the square roots 
of the corresponding numbers of the second. We may also re- 
mark that, the square of a number expressed by a single figure, will 
contain no figure of a higher denomination than tens. 

The numbers of the last line, 1, 4, 9, 16, &c., and all other 
numbers which can be produced by the multiplication of a num- 
ber by itself, are called perfect squares. 

It is obvious, that there are but nine perfect squares among all 
the numbers which can be expressed by one or two figures: the 
square roots of all other numbers expressed by one or two figures 

















ie 
OHAP. V.] EXTRACTION OF THE SQUARE ROOT OF NUMBERS. 109 


will be found between two whole numbers differing from: "each 
other by unity. Thus, the square root of 55, comprised between 
whe perfect squares 49 and 64, is greater than 7 and less than 8. 
Also, the square root of 91, comprised between the perfect squares 
31 and 100, is greater than 9 and less than 10. 

Every number may be regarded as made up of a certain num- 
yer of tens and a certain number of units. Thus 64 is made up 


of 6 tens and 4 units, and may be expressed under the form 
60 + 4 = 64. 

Now, if we represent the tens by a and the units by b, we 
‘shall have 
\) a +- b =60 + 4, 
jand (a + 6)? = (60 + 4), 
and consequently, 

a? + 2ab + B? — (60)? + 2 x 60 x 4+ (4)? = 4096. 
Hence, the square of a number composed of tens and units con- 
yains, the square of the tens, plus twice the product of the tens by 


| 


the units, plus the square of the units. 


117. If now, we make the units 1, 2, 3,4, &c., tens, by an- 
jaexing to each a cipher, we shall have, 

mio. 20, "80, 40," 50, ‘60, 70, 80, 90, 100; 
\and for their squares, 

100, 400, 900, 1600, 2500, 3600, 4900, 6400, 8100, 10000, 

| rom which we see that the square of one ten is 100, the square 
lof two tens, 400, &c.: and hence, the square of tens will contain no 
| figure of a less denomination than hundreds, nor of a higher name 
\‘han thousands. 


Let us now take any number, as 78, and square it. We have 
78 =70+48; 


He 


vhat is, equal to 7 tens, or 70, plus 8 units. 
Seven tens, or 70 squared - - (70)? = 4900 


twice the tens by the units is, 2 x 70 x 8 = 1120 
square of the units is, - - - (Sy72at 64 
hence, - - - - - , (78)? = 6084. 


Let us now reverse this process and find the square root of 


‘ 


110 ELEMENTS OF ALGEBRA. [CHAP. V 


Since this number is composed of more than two ita 
places of figures, its roots will contain more than one. 60 84 
But since it is less than 10000, which is the square |. 
of 100, the root will contain but two figures; that is, units and 
tens. 

Now, the square of the tens must be found in the two left- 
hand figures which we will separate from the other two, by 
placing a point over the the place of units, and another over 
the place of hundreds. These parts, of two figures each, are 
called periods. The part 60 is comprised between the two squares 
49 and 64, of which the roots are 7 and 8: hence, 7 ts the figure 
of the tens sought; and the required root is composed of 7 tens 
and a certain number of units. 


The figure 7 being found, we =e 
write it on the right of the given 60 84 | 78 
number, from which we separate saab 
it by a vertical line: then we | -7x* 2-148 1/1184 
subtract its square 49 from 60, 118 4 
which leaves a remainder of 11, 0 
to which we bring down the two 





next figures 84. The result of this operation is 1184, and this 
number is made up of twice the product of the tens by the units 
plus the square of the units. 

But since tens multiplied by units cannot give a product of a 
less name than tens, it follows that the last figure 4 can form no 
part of the double product of the tens by the units: this double 
product is therefore found in the part 118. 

Now, if we double the tens, which gives 14, and then divide 
118 by 14, the quotient 8 is the units’ figure of the root, or a 
figure greater than the units’ figure. ‘This quotient figure can 
never be too small, since the part 118 will be at least equal to 
twice the product of the tens by the units: but it may be too large; 
for, the 118 besides the double product of the tens by the units, 
may likewise contain tens arising from the square of the units. 


To ascertain if the quotient 8 expresses the units, we write the 
8 to the right of the 14, which gives 148, and then we multiply 
148 by 8. ‘Thus, we evidently form, Ist, the square of the units, 
and 2d, the double product of the tens by the units. This mul- 
tiplication being effected, gives for. a product 1184, a number equal 


iy , 
“HAP, V.] EXTRACTION OF THE SQUARE ROOT OF NUMBERS. 111 


o the result of the first operation. Having subtracted the prod- 
‘ct; we find the remainder equal to 0: hence 78 is the root 
equired. 

' Indeed, in the operations, we have merely subtracted from the 
iven number 6084, Ist, the square of 7 tens or of 70; 2d, twice 
‘he product of 70 by 8; and 3d, the square of 8: that is, the 
‘hree parts which enter into the composition of the square of 78. 


| Remarx.—The operations in the last example have been per- 
‘ormed on but two periods. It is plain, however, that the same 
“easoning is equally applicable to larger numbers; for, by chan- 
zing the order of the units, we do not change the relation in 
which they stand to each other. 
_ Thus, in the number 60 84 95, the two periods 60 84, have 
che same ‘relation to each other, as in the number 6084; and 
hence, the methods pursued in the last example are equally ap- 
plicable to larger numbers. 

Hence, for the extraction of the square root of numbers, we 
have the following 


| 


RULE. 


I. Separate the given number into periods of two figures each, be- 
ginning at the right hand: the period on the left will often contain 
but one figure. 

Il. Find the greatest square in the first period on the left, and 
‘place its root on the right after the manner of a quotient in division. 
‘Subtract the square of the root from the first period, and to the 
remainder bring down the second period for a dividend. 

III. Double the root already found and place it on the left for a 
divisor. Seek how many times the divisor is contained in the divi- 
‘dend, exclusive of the right-hand figure, and place the figure in the 
‘root and also at the right of the divisor. 

IV. Multiply the divisor thus augmented, by the last figure of 
‘the-root, and subtract the product from the dividend, and to the re- 
mainder bring down the next period for a new dividend. 

| V. Double the whole root already found, for a new divisor, and 
‘continue the operation as before, until all the periods are brought 
down. 


I. Remarx.—lf, after all the periods are brought down, there is 
‘no remainder, the proposed number is a perfect square. But if 


me 
112 ELEMENTS OF ALGEBRA. [CHAP. V 


there is a remainder, we have only found the root of the greatest 
perfect square contained in the given number, or the entire pari 
of the root sought. 

For example, if it were required to extract the square root of 
168, we should find 12 for the entire part of the root and a re- 
mainder of 24, which shows that 168 is not a perfect square. 
But is the square of 12 the) greatest perfect square contained in 
168? That is, is 12 the entire part of the root? To prove this, 
we will first show that, the difference between the squares of two 
consecutive numbers, is equal to twice the less number augmented 


by unity. : 
Let a = the less number, 
and a-+1 = the greater. 
Then (a+ 1? =@+2a+41 
and C3 john ag 
Their difference is — 2a +1 as enunciated. . 


Hence, the entire part of the root cannot be augmented by 1, un 
less the remainder is equal to, or exceeds twice the root found, plus 
unity. 

But, 12x 2+1=25; and since the remainder 24 is less 
than 25, it follows that 12 cannot be augmented by a number as 
great as unity: hence, it is the entire part of the root. 

The principle demonstrated above, may be readily applied im 
finding the squares of consecutive numbers. 

If the numbers are large, it will be much easier to apply the 
above principle than to square the numbers separately. 


For example, if we have (651)? = 423801; 
and wish to find the square of 652, we. have 
(651)? = 423801 





+2x 651 = 1302 

re Be Meo Lon 
and (652)? = 425104. 
Also, (652)? = 425104 
Di tot, 2 Oa eeemmare Vere 
+L acs 1 


(653)? — 426409. 


a 














‘CHAP. V.] EXTRACTION OF THE SQUARE ROOT OF NUMBERS. 1t3 


Mf. Remarx.—The number of figures in the root will always 
he equal to the number of periods into which the given number 
is separated. 


EXAMPLES. 


. To find the square root of 7225. 

. To find the square root of 17689. 

. To find the square root of 994009. 

. To find the square root of 85678973. 
. To find the square root of 67812675. 


own & 


Of Incommensurable Numbers. 


° . e ° 
_ 118. If a number is not a perfect square, its square root is 
| said to be tncommensurable, or irrational, because it cannot be ex- 


lessed in terms of the numerical unit. Thus, Jf 2 ; By : Wi : 
jare incommensurable numbers. They are also sometimes called 
\radicals or surds. 

_ Two or more numbers are said to be prime with respect to 
feach other, when there is no whole number except unity which 
\will divide each of them without a remainder. Thus, the num- 
jers 3 and 5 are prime with respect to each other; and so also 
fue '4 and 7 and 9. 

| In order to prove that the root of an imperfect power cannot 
le expressed by exact parts of unity, we must first show that, 
i Every number P, which will exactly divide the product A x B 
if two numbers, and which is prime with one of them, will divide 
he other. | 

Let us suppose that P will not divide A, and that A is greater 
than P. 

| Let us now find the greatest common divisor of A and P. If 
. ve represent the entire quotients by QQ?) O74: idee. andthe 
‘emainders, respectively, by R, R’, R”, &c.; we shall have 








is i hence, A= PQ+R, 

i i hence, ° P = RQ’ + R’, 

e i hence, R= R’/Q” + R”, 
*: bee hence, | Ries RC OMe aR, 


8 


List ELEMENTS OF ALGEBRA. [CHAP. V. 


Now, since he emainders R, R’, R’, &c., constantly dimin- 
ish, if the division be continued sufficiently far, we shall obtain 
a remainder equal to unity ; for the remainder cannot be 0, since 
by hypothesis A and P are prime with each other. Hence, we 
have the following equations : 


A=P.Q +R 
P=RQ +P 
R =R’Q’ +R” 
R/ inion R“O“ + RR” 


Multiplying the first of these equations by B, and dividing by 
P, we have 





AB ue 
i ee ; 
But, by hypothesis, p is an entire number, and since E 


and Q are entire numbers, the czoduct BQ is an entire number 


Hence, it follows that is an entire number. 





If we multiply the second of the above equations by B, anc 
divide by P, we have 
BRQ’ 
B — he =| a 


is an entire number 





But we have already shown, that 
‘aa ; BR’ 
p is an entire number. This being the case, “7 
must also be an entire number. If the operation be continue 
until the number which multiplies B becomes 1, we shall hav 
Bx 1 

a 
vide B. 

In the operations above we have mee A> P; butt 
P >A, we should first divide P by A | 


Hence, if a number P will exactly divide the product of two num 
bers, and is prime with one of them, it will divide the other. 


hence 








equal to an entire number, which proves that P will di 











HAP. V.] EXTRACTION OF THE SQUARE ROOT OF FRACTIONS. cS 


We see from what has preceded that, of P is prime with re- 
sect to any number as a, it will also be prime with respect to a? 
‘nd the higher powers of a. 

For, if P will divide a? = ax a, it must divide one of the fac- 
ms @ or a. But this would be contrary to the supposition; hence, 
cannot divide a*. In the same way it may be proved that it 
mnot divide the higher powers of a. 

We will now show that the square root of an imperfect square 
mnot be expressed by a fractional number. 
_Let ¢ be an imperfect square. Then if its exact root can be 
xpressed by a fractional number, we can assume 


Vea, 


| Er hE et we, 
‘, which the fraction |Z (is in its lowest terms: that is, in 


vhich @ and 6 are prime with respect to each other. 
| Now, if we square both members of the equation, we have 
) 2 


| (OES ee 
i, * $2’ 





‘ which ¢ is an entire number: and hence, if the equation is 
jue, a? must be divisible by 6?. 
‘ But if a? is divisible by 6?, the product a x a = a?, must be 


Extraction of the Square Root of Fractions. 


| 119. Since the second power of a fraction is obtained by squar- 
1g the numerator and denominator separately, it follows that the 
|uare root of a fraction will be equal to the square root of the 
}imerator divided by the square root of the denominator. 


| 2 
. For example, ve = > 


since . m4 
b 





116 ELEMENTS OF ALGEBRA. (CHAP. V, 


But if the numerator and the denominator are not both perfect 
squares, the root of the fraction cannot be exactly found. We 
can, however, easily find the exact root to within less than one 
of the equal parts of the fraction. For this purpose, 


Multiply both terms of the fraction by the denominator—this makes 
the denominator a perfect square. Then extract the square root of 
the perfect square nearest the value of the numerator, and place the 
root of the denominator under it—this fraction will be the appromi- 
mate root. 


3 


Thus, if it be required to extract the square root of e 


we 
. : ; 15 

multiply both terms by 5, which gives ae the square nearest 
‘ Sone : , 

15 is 16: hence, Fri the required root, and is exact to with- 


g 1 
in less than —. 


120. We may, by a similar method, determine, approximatively, 
the roots of whole numbers which are not perfect squares. Let 
it be required, for example, to determine the square root of an 


3 1 ; 
entire number a, nearer than the fraction —; that is to say, to 
n 
find a number which shall differ from the exact root of a, by a 
yi l 
quantity less than —. It may be observed that, 
n 


an2 


a=. 
n2 


If we designate by r the entire part of the root of an?, the 
number an? will then be comprised between r? and (r + 1)?; and 


1} aos #2 r-y- LY? 
— will be comprised between — > and hog is and conse 
n n® n? 
quently the true root of a is comprised between 
ie rainy | 
\/ andl 4) oe 


n 
r+1 
ca 





that is, between and But the difference between 


; 

n 
, l 
these numbers is — 
n 


Cee 
hence = will represent the square root 


} 
‘DHAP. V.] EXTRACTION OF THE SQUARE ROOT OF FRACTIONS. 117 
] 
tof a within less than the fraction —. Hence to obtain the root: 


Multiply the given number by the square of the denominator of 
the fraction which determines the degree of approximation: then ex- 
‘Tact the square root of the product to the nearest unit, and divide 
las root by the denominator of the fraction. 


| 1. Suppose, for example, it were required to extract the square 


: 


; rhe 1 
oot of 59, to within less than et 


First, (12)? = 144; and 144 x 59 = 8496. 
| Now, the square root of 8496 to the nearest unit, is 92: hence 


12 Seal AS 1 
= 7,5, which is true to within less than Ip 


2. To find the +/11 to within less than = 


4 
A ° 3, 
ns 15 









| 3 To find the 4/223 to within less than sa 


37 
i Ans. 14—. 
msi 14s 
| 121. The manner of determining the approximate root in deci- 
jjals, is a consequence of the preceding rule. 


To obtain the square root of an entire number within io’ 
‘oe oe 
ia’ 7000" &c., it is only necessary according to the preceding 


jile, to multiply the proposed number by (10)?, (100)?, (1000)2; 
|, which is the same thing, 


: Annex to the number, two, four, six, &c., ciphers: then extract 
e root of the product to the nearest unit, and divide this root by 
J, 100, 1000, &c., which ts effected by pointing off one, two, three, 
c., decimal places from the right hand. 


EXAMPLES. 


1. To find the square root of 7 to within 


118 ELEMENTS OF ALGEBRA. [CHAP. V 


Having multiplied by (100)?, that is, 70000| 2.64 
having annexed four ciphers to the right . 4 | 


hand of 7, it becomes 70000, whose 46 300 


root extracted to the nearest unit, is 276 

4, which being divided by 100 gives ae 
dentrepaiies Wet Mase 524 | 2400 
2.64 for the answer, which is true to 

1 2096 
within less than 100" 304 Rem 
pa nie 1 
2. Find the AA 29 to within —— Ans. 5.38. 


100° 


3. Find the ~ 227 to within Ans. 15.0665. 


1 
10000° 

Remarx.—The number of ciphers to be annexed to the whol, 
number, is always double the number of decimal places require 
to be found in the root. 


122. The manner of extracting the square root of decimal frac 
tions is deduced immediately from the preceding article. 

Let us take for example the number 3.425. This fraction i 
3425 
1000° 
denominator may be made such without altering the value of th 


equivalent to Now 1000 is not a perfect square, but th 


fraction, by multiplying both the terms by 10; this gives Todt 


4 
r aa Then extracting the square root of 34250 to th 


nearest unit, we find 185; hence or 1.85 is the requir 










185 

100 
root to within et 
100 

If greater exactness be required, it will be necessary to add 

the number 3.4250 so many ciphers as shall make the perio 

of decimals equal to the number of decimal places to be found 

the root. Hence, to extract the square root of a decimal fractio 


Annex ciphers to the proposed number until the number of de 
mal places shall be equal to double the number required in the ro 
Then extract the root to the nearest unit, and point off from t 
right hand the required number of decimal places. 


SHAP. V.] SQUARE ROOT OF ALGEBRAIC QUANTITIES. 119 


EXAMPLES. 
1. Find the V 3271.4707 to within .0Ol. Ans. 57.19. 
2. Find the + 31.027 to within .01. Ans. 5.57. 


3. Find the 0.01001 to within .00001. Ans. 0.10004. 


123. Finally, if it be required to find the square root of a vul 
zar fraction in terms of decimals: 


Change the vulgar fraction into a decimal and continue the di- 
yision until the number of decimal places is double the number re- 
quired in the root. Then extract the root of the decimal by the last 
rule. 


EXAMPLES. 


11 Mes F 
1. Extract the square root of cm to within .001. This num- 


per, reduced to decimals, is 0.785714 to within 0.000001. The 
root of 0.785714 to the nearest unit, is .886: hence 0.886 is the 


iE oes 
root of a to within .001. 


2. Find the \/ 2 to within 0.0001. Ans. 1.6931. 


Extraction of the Square Root of Algebraic Quantities. 


124. Let us first consider the case of a monomial. In order 
to discover the process for extracting the square root, let us see 
how the square of a monomial is formed. 

By the rule for the multiplication of monomials (Art. 41), we 
have 

(5ab%c)? = 5a*b3c X 5a*bc = 25atb%e? ; 
that is, in order to square a monomial, it is necessary to square 
its co-efficient, and double the exponent of each letter. Hence, te 
find the square root of a monomial, 


Ist. Extract the square root of the co-efficient and divide the ex- 
ponent of each letter hy two. 2d. To the root of the co-efficient 
annex each letter with its new exponent, and the result unll be the 
required root. 


a 


120 ELEMENTS OF ALGEBRA. [CHAP. V 


Thus, a/ 64a°b+ = 8a°b?; for, 8a3b? x 8a%b? = 64aD+, 

and,.. 4/ 625a7b®cS§ — 25ab‘tc? ;;_ for, (2bah*e?)* = G200-0'c". 

125. From the preceding rule, it follows, that, when a monomial 
is a perfect square, its numerical co-efficient is a perfect square, and 
the exponent of every letter an even number. ‘Thus, 25a*b? is a 
perfect square, but 98ab* is not a perfect square; for, 98 is not 
a perfect square, and a is affected with an uneven exponent. 

An imperfect square is introduced into the calculus by affecting 


it with the radical sign ¥/ __, and written thus, tf 98ab!. Quan- 
tities of this kind are called radical quantities, or irrational quan- 
tities, or simply radicals of the second degree. 

These expressions may sometimes be simplified. For, by the 
definition of the square root, we have 


Va xVa =(fapP =a, 
vab x Vab =(7ab)? = ab, 
Jabe x Vabe =(Vabe)? = abe, 
Vabed x abcd = (abcd)? = Whed ; 
and the same would be true for any number of factors. 
Again, 
(Va.Vb.Ve.Vdp=(Vap(y dP. (ep. (dp = ated 
by the rule for multiplying monomials (Art. 41). 
Now, since, (/ abcd)? = abcd, 
and, (af at ABD OF / d)? = abed; 
it follows, that the quantities themselves are equal: hence, 
Vabed =Ja.J/b.Vc.V/d; that is, 
The square root of the product of two or more factors is equal to 


the product of the square roots of those factors. 


This being proved, we can write 


J 98ab* — /4904 x 2a — 4904 x o/2a. 
But, a/ 49b* — 722: 
hence, of DSabtas= ‘75+ o/ 2a. 


HAP. V.] SQUARE ROOT OF ALGEBRAIC QUANTITIES. 121 


- In like manner, 


o/ 45a2b3e2d = +/9a2b%c? x 5bd = Babe »/5bd. 


o/ 864.a2b%cH} mans 144a70*e\~ Ghe'=. 12ab-e o/ 6be. 
| The quantity which stands without the radical sign is called 
the co-efficient of the radical. Thus, 7b?, 3abc, and 12ab?c°, are 
o-efficients of the radicals. 

In general, to simplify a radical of the second degree: 


; 
» Dwide the quantity under the radical sign by the smallest mono- 
‘ual, with reference to its co-efficients and exponents, that will give 
‘or a quotient a perfect square. Then, extract the root of the per- 
‘eet square and place it without the radical sign, under which, write 
ihe monomial used as a divisor. 





EXAMPLES. 
. To reduce ad 75a3be to its simplest form. 


. To reduce 4/128d%a°d? to its simplest form. 


, 


1 
2 
3. To reduce Hy 32a%8e to its simplest form. 

4. To reduce o/256a2b*c8 to its simplest form. 
5. To reduce af 1024a%7c> to its simplest form. 
6. To reduce af 728a7bec8d to re simplest form. 


| 126. Since like signs in both the factors give a plus sign m 
ie product, the square of —a, as well as that of +a, will be 
2: hence, the root of a? is either +a or —a. Also, the square 
oot of 25a2b* is either + 5ab? or — 5ab?. Whence we may con- 
lude, that if a monomial is positive, its square root may be af 
‘ either with the sign + or —; 

| 


thus, wavaeiqe <= 3a?, 


mw, + 3a? or — 3a, squared, gives 9a*. The double sign + 
|rith which the root is affected, is read plus or minus. 

If the proposed monomial were negative, it would have no square 
90t, since it has just been shown that the square of every quan- 
ty, whether positive or negative, is essentially positive. There- 


ore, 
4/9, af — 4a, + — 8a’, 


!re algebraic symbols which indicate operations that cannot be 


6 





122 ELEMENTS OF ALGEBRA. [CHAP. V,} 


performed. They are called imaginary quantities, or rather, 7 | 
aginary expressions, and are frequently met with in the resolution| 
of equations of the second degree. 


127. Let us now examine the law of formation of the square) 
of a polynomial; for, from this law, the rule is deduced for ex- 
tracting the square root. 

It has already been shown (Art. 46), that, 

(a + 5)? = a? + 2ab + 67; that is, 

The square of a binomial is equal to the square of the first term 
plus twice the product of the first term by the second, plus the 
square of the second. | 






The square of a polynomial, is the product arising from mult. 
plying the polynomial by itself once: hence, the first term of the 
product, arranged with reference to a particular letter, is the 
square of the first term of the polynomial arranged with reference 
to the same letter. Therefore, the square root of the first term 
of such a product will be the first term of the required root. 


128. Let us now extract the square root of the polynomial 
28a + 49at + 4a® + 9 + 42a? + 1203, 
which arranged with reference to the letter a, becomes, 
4a’ + 2805 + 49at+ 1203+ 42a?+9| 2a3+ T7a?+3 
















4a? 4a3+ Ta? 
R = 28a5 + 49a4-+ 1203+ 42a? + 9 7a? 
28a? + A0a" 2805+ 49at = (2r+ 1) 
R’ = - 12a3 + 42a? +9] 4a3+ 140?+3 
1243 + 420249 3 
Rs re 0 0 0] 1242+ 42a?+ 9 = (2n+ rr’) 





Now, since the square root of 4a® is 2a°, it follows that 2Qa' 
is the first term of the required root. ‘Designate this term by 1' 
and the following terms of the root, arranged with reference to @ 
byangr's ri éce! 

Now, if we denote the given polynomial by N, we shall have! 

=—(rtrtr’+ rr” + &e.;)? 
or, if we designate all the terms of the root, after the first, by § 
IN == (++ 5)? er ors 4-84 
= 2 4 Or (+r + 7” + &e.) + 8. 


-GHAP. V.] SQUARE ROOT OF ALGEBRAIC QUANTITIES. 123 





If now we subtract r? = 4a5, from N, and designate the re- 
mainder by R, we shall have 


R= N — 408 = 27 (Ph +r’ 4+ 774+ &c.) + s?; 


_in which the first term 2r7’ will contain a to a higher power 
than either of the following terms. Hence, if the first term of 
the first remainder be divided by twice the first term of the root, 
| the quotient will be the second term of the root. 





, If now, we place pil beh win 


‘and designate the remaining terms of the root, 1”, 7”, &c., by s’, 
_we shall have 
N = (n + s’)? = n? + 2ns’ + 8; and 

. R’ = N — n? = (2r 4 27’) (77 + 1” + &e.) + 8/2; 

}in which, if we perform the multiplications indicated in the sec- 
»ond member, the’ term 277” will contain a higher power of a than 
‘either of the following terms. Hence, if the first term of the 
second remainder be divided hy twice the first term of the root, the 

quotient will he the third term of the root. 


If we make 
r+-nreot+r=—n, and r’+7rV¥4+ &e.=s%, 
we shall have 
N = (n’ +5")? = n? + 2n’s’4+ ’; and 
R’ =N—2n?=2(r+r4+r)(r’%+ 77+ &.) + 82; 


from which we see, that the first term of any remainder, divided 
by twice the first term of the root, will give a new term of the re- 
quired root. 

It should be observed, that instead of subtracting n? from the 
given polynomial, in order to find the second remainder, that that 
remainder could be found by subtracting (27+ 7’)r’ from the 
first remainder. So the third remainder may be found by sub- 
tracting (2n+77”)r” from the second, and similarly for the re- 
mainders which follow. | 

In the example above, the third remainder is equal to zero, 
and hence the given polynomial has an exact root. 

Hence, for the extraction of the square root of a polynomial, 

we have the following 


- 


124 ELEMENTS OF ALGEBRA. [CHAP. V 


RULE. 


I. Arrange the polynomial with reference to one of its letters, 
and then extract the square root of the first term, which will give 
the first term of the root. Subtract the square of this term from 
the given polynomial. 

Il. Divide the first term of the remainder by twice the first term 
of the root, and the quotient will be the second term of the ‘root. 

III. From the first remainder subtract the product of twice the 
first term of the root plus the second term, by the second term. 

IV. Divide the first term of the second remainder by twice the 
first term of the root, and the quotient will be the third term of 
the root. 

V. From the second remainder subtract the product of twice the 
first and second terms of the root, plus the third term by the third 
term, and the result will be the third remainder, from which the 
fourth term of the root may, be found; and proceed in a similar 
manner for the remaining ters of the root. 


EXAMPLES. 
1. Extract the square root of the polynomial 
49a7b? — 24ab3 + 25a* — 30a5d + 1654. 
First arrange it with reference to the letter a. 


25at — 30a3b + 49a2b? — 24ab3 + 165+ 5a* — 3ab + 45? 





25a* 10a? — 3ab 
R =— 3003 + 49a2b? — 24ab3 + 1654 — 3ab 
— 30a3b + 9a2h? 30a3b + 9a2b? 
R’ = + 40a2b2 — 24ab3 + 1654 | 10a? — 6ub + 462 
+ 40a2b? — 24ab3 + 1654 4b? 
R’“= - - - a - - 40a2b2 — 24ab3 + 1654 


2. Find the square root of 
at + 4a3x + 6a?x? + 4ax3 + of, 
3. Find the square root of 
at — 2a3n + 3a2x? — 2ax3 + at, 
4. Find the square root of 
4x6 + 1245 + 5at — 2x3 + 7x? — Qa + 1, 


\CHAP. V.] SQUARE ROOT OF ALGEBRAIC QUANTITIES. 125 


5. Find the square root of 


9at — 12a°b + 280262 — 16ab3 + 1684. 
6. Find the square root of 


(25atb? — 400% + 760?b2c? — 48ab%3 + 3682ct — 30a%be + 24a3be? 
— 36a7bc3 + 9atc?. 


129. We will conclude this subject with the following remarks: 
Ist. A binomial can never be a perfect square. For, its root 
‘cannot be a monomial, since the square of a monomial will be a 
|monomial; nor can the root be a polynomial, since the square of 
whe simplest polynomial, viz., a binomial, will contain at least 
‘three terms. Thus, an expression of the form 





lar mel 


k 


jcan never be a perfect square. 


: 2d. A trinomial, however, may be a perfect square. If so, 
/ when arranged, its two extreme ternis must be squares, and the 
|middle term double the product of the square roots of the other 
‘wo. Therefore, to obtain the square root of a trinomial, when it 
|s a perfect square, ; 


| Extract the roots of the two extreme terms, and give these roots 
he same or contrary signs, according as the middle term is posi 
‘we or negative. To verify it, see if the double product of the two 
|"oots is equal to the middle term of the trinomial. Thus, 





| 9a* — 48atb? + 64a7b4 is a perfect square, 

s for, +/..9a°)— 3a3, and a/ 64a2b+ — — 8ab?, 
and also, 2 x 3a? x — 8ab? — — 48a‘tb?, the middle term. 
But 4a? + 14ab + 93? 


8 not a perfect square: for, although 4a? and + 90? are the 
| quares of 2a and 3), yet 2 x 2a x 3d is not equal to 14ab. 


3d. When, in extracting the square root of a polynomial, the 
irst term of any one of the remainders is not exactly divisible 
ly twice the first term of the root, we may conclude that the 
iroposed polynomial is not a perfect square. This is an evident 
onsequence of the course of reasoning, from which the general 
ule for extracting the square root was deduced. 





126 ELEMENTS OF ALGEBRA. [CHAP. V. 


4th. When the polynomial is not a perfect square, the expres- 
sion for its square root may sometimes be simplified. 
Take, for example, the expression 
Jab 4a E 4aB. 
The quantity under the radical is not a perfect square: but i 
can be put under the form 


ab (a? + 4ab + 467). 


Now, the factor within the parenthesis is evidently the square of 
a+ 2b, whence we have . 


o/a3b + 402d? + 4ab3 = (a + 2b) ab. 


Of the Calculus of Radicals of the Second Degree. 


130. A radical quantity is the indicated root of an imperfect 
power. If the root indicated is the square root, the expression is 
called a radical of the second degree. ‘Thus, 


V2 8-4) Ded Ay ae 
are radicals of the second degree. 


131. Two radicals of the second degree are similar, when the 
aes under the radical sign are the same in both. Thus, 


BNE b and 5e +/ b are similar radicals; and so also, are 9+/ 2 


and 7+ 2 


Addition and Subtraction. 


132. In order to add or subtract similar radicals, add or sub- 
tract their co-efficients, and to the sum or difference annex the com- 
mon radical. 


Thus, 3a/ b + 5c b = (3a + 5c) J b; 
and 3a./b —5e b =(3a—5c) 7 b. 


In like manner, 


72a +3 V/2a =(74+ 3) V2a= 10 2a; 
and Va/ 2a 84/20 =\(7 = ayRfage, aay oe 


Two radicals, which do not appear to be similar, may become 
so by simplification (Art. 125). 


HAP. V.] RADICALS OF THE SECOND DEGREE. 4127 


|, When the radicals are not similar, their addition or subtraction 
Ie only be indicated. Thus, to add 3+ 6 to 5+/ a, we write 


Big 34) 8» 


ee 


Hieyloan> Aiki b)? = ab; also 

that, (o/ ab b)? = ab; hence, 
(fa x a5: = (4/ ab)? ; and consequently, 
Ja x Jb = Va: that is 


The product of two radicals of the second degree is equal to the 
| yware root of the product of the quantities under the radical signs 





When there are co-efficients, we first multiply them together, and 
bite the product before the radical sign. Thus 


Sab x 4 20a = 12 ¥/100a2 = 120a / b 

2avV/ be X 38a+v be = 6a? af b2c2 = 6a*be. 

2aVa?+ hb? x — 3a a? + b? = — 6a? (a? + 52). 

Diwision. ° 

| 134. To divide one radical by another, let us observe that 
war? (fap a 
( ) — =— also, 

/ Vf b (Vv 6) 








3 a 
The quotient of two radicals is equal to the square root of the 
swotient of the quantities under the radical signs. 





128 ELEMENTS OF ALGEBRA. FCHAP. V. 


‘When there are co-eflicients, write their quotient as the co-effi. 
cient of the radicals. 


For example, 
bab 2207 c= Ley fh 
And 12ac \/ 6be ~ 4c 4/ 2b 3aV\/ as = 3a J 3c. 


135. There are two transformations of frequent use in finding 
the numerical values of radicals. 

The first .transformation consists in passing the co-efficient of 
a radical under the radical sign. Take, for example, the ex 


pression 3a Ae 5b. By applyimg the rules for the multiplication of 
radicals we may write, 


a /5b = /9a2 x V/5b = Vf 9a? x 5b = 450%. 


Therefore, the co-efficient of a radical may be passed under the rad 
ical sign, as a factor, by squaring it. 

The principal use of this transformation, is to find an approxi- 
mate value of any radical, which shall differ from its true value, 
by less than unity. 

For example, take the expression 64/13 13. Now, as 13 is no 
a perfect square, we can only find an approximate value for its 
square root; and when this approximate value is multiplied by 6 


the product will differ materially from the true value of 6 13. 
But if we write, 


6/13 = /62 x 13 = 1/36 x 13 = 1/468, 
we find that the square root of 468 is the whole number 21 
plus an irrational number less than unity. Hence, 
6 4/13 = 21 plus an irrational number less than 1. 
In a similar manner we may find, 


12+“ 7 =31 plus an irrational, number less than 1. 


136. Having given an ae of the form, 


p+ ee q 
in which @ and p are any wee whatever, and q not a per- 
fect square, it is the object of the second transformation to ren- 
der the denominator a rational quantity. 


" 
“MAP. V.]J RADICALS OF THE SECOND DEGREE. 129 


| This object is attained by multiplying both terms of the frac- 
on by p—-+vq, when the denominator is p+ q, and by 
+ /q, when the denominator is p—V4q: g; and recollecting 


rat the sum of two quantities, multiplied by their difference, is 
qual to the difference of their squares: hence, 


a a(p—V 9) _ a(p—V 9) _p—av gq 
bt+Va (PtV a0 We iP a aa 


fee a(p+vq) _@ptv9)_wptavg 
, (ey <7 p> —4q p?—4q i 
in which the denominators are rational. 


As an example to illustrate the utility of this method of ap- 
‘roximation, let it Ps required to find the approximate value of 




















y 

‘ae expression TF We write 

‘ 

Me 8 i/o) 2h 7 V8 
ey ae ee ie (halt 

' But, 7/5 =J/49 x 5 = 245 — 15+ an irrational 


! 





umber less than unity. Therefore, 
7 ; Al 21 + 1 15 + irr. number er <1 1 
leat, ./5 4 


ence, 9 differs from the true value by less than one fourth, hen 





(spain 
4 


If we wish a more exact value for this expression, extract the 
quare root of 245 to a certain number of decimal places, add 21 
0 this root, and divide the result by 4. | 


For another example, take 
DA, 
fiu+ 73 


nd find its value to. within 0.01. 
We have, 


fs Wifi J _ 18-78 
Sa /3 ° 8 8 : 





130 ELEMENTS OF ALGEBRA. [CHAP. V. 


Now, 74/55 = »/55 x 49 = »/2695 = 51.91, within 0.01, 


and 4/15 4/15 x 49 = 4/795 = 2a = 
therefore, 
775 O19) — 27.11 24.80 


ee alee ens ven = —— = 3.10. 
Viul+ 73 8 8 


Hence, we have 3.10 for the required result. This is exact te 


1 
ithin ——., 
within => 
By a similar process, it will be found that 
aye ey A 


p Pe cel i 2 Powis aay l Sats it to within 0.001. 

Remarx.—The value of expressions similar to those above, may 
be calculated by approximating to the value of each of the rad- 
icals which enter the numerator and denominator. But as the 
value of the denominator would not be exact, we could not deter- 
mine the degree of approximation which would be obtained, where- 
as by the method just indicated, the denominator becomes rational, 
and we always know to what degree the approximation is made. 


Examples in the Calculus of Radicals. 


We make the reductions in the examples which follow, accord- 
ing to the methods indicated in Art. 125; though, it is sometimes 
necessary to multiply the quantity under the radical sign, instead 
of dividing it. 

1. Reduce 1/125 to its simplest form. 


We first seek the largest perfect square that will exactly di- 
vide 125. We try, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, and 
144. We find that 25 is the only one that will give an exact 


quotient: hence, 
J 125 = f25x5=5/5. 


50 
. Reuuc \/ bal te its si A 
ie”. euuce ia? to its simplest form 


We observe net 25 will divide the numerator, and hence 


oe aa X <= [ 2 
7 147° 





HAP. V.] EXAMPLES IN THE CALCULUS OF RADICALS. 13] 


‘Since there is no perfect square which will divide 147, we 
ust see if we can multiply it by any number which will give 


perfect square for a product. Multiplying by 2 we have 284, 

3 i 
hich is not a perfect square. ‘Then trying 3, we find the prod- 
*t 441, whose square root is 21. Hence, we have 


[ 2 f 2x3 f cs 
cya 147 x37 4a e mage 


3. Reduce / 98a2x 98a?x to its most simple form. 


Ans. Ta a/ 22. 


4, Reduce af (a? — ax”) to its most simple form. 


5. Required the sun of 72 and vy 128. 


Ans. 142. 
6. Required the sum of 1/27 and 147. 
Ans. 10+ 3. 
2 Ve 
7 Required the sum of fie and 50" 
RF a 
Ans. — 
ns =v 6 


8. Required the sum of 2/a’b and 3+/ 64bz%. 





9. Required the sum of 9+/243 and 10 4/363. 


3 
. 10. Required the difference of = and Ve 


14 “ 
Ans. 45 15.. = 
th. Required the product of 5 A ‘8 and 37/5. 
Ans. 30 4/10. 


' 1 
12. Required the product of Bf and A 


Ans. av 35 
. Divide 6/10 by 3/5. 
. What is the sum of +/48ab? + b /75a. 


. What is the sum of +/ 18a5b? + +/ 50a%d3. 
Ans. (3a2b + 5ab) +/ 2ab. 


132 ELEMENTS OF ALGEBRA. (CHAP. VI, 


CHAPTER VI. 
EQUATIONS OF THE SECOND DEGREE. 


137. An Equation of the second degree is one in which the 
greatest exponent of the unknown quantity is equal to 2. If the 
equation contains two unknown quantities, it is of the second de- 
gree when the greatest sum of the exponents with which the 
unknown quantities are affected, in any term, is equal to 2. 


138. Equations of the second degree are divided into two 
classes. | 
1st. Equations which involve only the square of the unknown 
quantity and known terms. These are called incomplete equations. 
2d. Equations which involve the first and second powers of the 
unknown quantity and known terms. These are called complete 
equations. 
Thus, et 2a?7—-5=7 
and 5x2? — 32? —4 =a, 
are incomplete equations ; and 
32? — 5a — 3a? +a=—) 
2a* — 8a? — 2 —~c=d, 
are complete equations. 


Of Incomplete Equations. 


139. The following is the most general form of an incomplete 
equation : viz., 


If we reduce this to an equation containing only entire terms, 
we have, 


6aca? — 6ba? — 7c = Ged: 


‘SHARP. VI.] EQUATIONS OF THE SECOND DEGREE. 133 


hence, ax” (6ac — 6b) = 6ed + 7c, 
6cd + Y Ley 

" Hee 
Bod, 2 —~ Gace — 6b 


vy substituting m, for the known terms which compose the sec- 
md member. .Hence, every incomplete equation cun be reduced to 
m equation involving but two terms, of the form 


i alt ’ ' 
and from this circumstance, the incomplete equations are often 
called, equations involving two terms. 


There is no difficulty in resolving equations of this form; for, 


i 


i If m is a perfect square, the exact value of # can be found 
by extracting its square root, and the value will be expressed 
either algebraically or in numbers. 

| If m is an algebraic quantity, and not a perfect square, it must 
be reduced to its simplest form by ‘the rules for reducing radi- 
‘cals of the second degree. If m is a number, and not a perfect 
‘square, its square root must be determined approximatively by the 
‘rules already given. 

' But the square of any number is +, whether -the number it- 


self have the + or — sign: hence, it follows that 


(+ J/my=m, and (— /m)?=m; 


and therefore, the unknown quantity a is susceptible of two dis- 
tinct values, viz. 


x=++HfYm, and «=—vVm; 
and either of these values being substituted for x will verify the 
given equation. For, 


a ode +J/mx+ lm =m; 
and aXxe= ea —/mx—Ym=n. 


The root of an equation is any expression which, being substi- 
‘tuted for the unknown quantity, will satisfy the equation ; that is, 
“make the two members equal to each other. Hence, every incom- 
plete equation of the second degree has two roots which are numeri- 
cally equal to each other; one having the sign plus, and the other 
the sign minus. 


we have elm. 


134 ELEMENTS OF ALGEBRA. (CHAP. VI 


1. Let us take, as an example, the equation 
1 5 cf 299 
Se pair Weer 
go tae a 
which, by making the terms entire, becomes 

8x? — 72 + 10x? = 7 — 24a? + 299, 
and by transposing and reducing 

420° = 878). and v2 == —— = 9; 


hence, w2=++/9 =+3; and ema 9 = —3. 
2. As a second example, let us take the equation 

om? mal, 
Dividing by 3 and extracting the square root, we have 


5 1 
Se PEPY fap 


in which the values must be determined approximatively. - 
3. What are the values of a in the equation 
11 (a? — 4) = 5 (a? + 2). Ans) = sb 3: 
4. What are the values of x in the equation 


a2 2 
m? — x m 
Vniaa Ans. 2= + ———_. 
x 1 + n? 





Of Complete Equations of the Second Degree. 


140. The most general complete equation of the second degreé 
can be expressed under the form 
ax? + bx — 3x —c=d; 
which may be put under the form 
ax? +(b—3)xa=d+e; 
and by dividing by the co-efficient of «, we have 
b—'3 d+e 


x? + ——_-¢= e 
a a 





If now, we substitute 2p for the co-efficient of x, and represen' 
the value of the second member by gq, we shall have, 


a -+- 2px = gq. 









HAP. VI.] EQUATIONS OF THE SECOND DEGREE. 135 


- The reduction to the above form is made: 


Ist. By transposing all the terms involving a? and @ to the 
‘irst member, and all the known terms to the second. 


2d. If the term involving x? should be negative, the signs of 
; ll the terms of the equation must be changed to render it posi- 
‘ive, and then divide both members by the co-efficient of «?. Hence, 
very complete equation of the second degree can be reduced to an 
|quation involving but three terms, and of the above form. The 
|juantity g is called the absolute term. 

If we compare the first member of the equation 


| ee -s 2pn== g, 
vith the square of a binomial ‘ 
I (~7 + a)? = a? + 2ax+ a? 

ve see that it needs but the square of p to render it a perfect 
square. If then, p* be added to the first member, it will become 
. perfect square; but in order to preserve the equality of the 
members, p? must also be added to the second member. Making 
vhese additions, we have 


| es a i a 
his is called completing the square, which is done by adding the 


square of half the co-efficient of « to both members of the equa- 
lon. 






Now, if we extract the square root of both members, we have, 


t+p=++7q¢-+ p’, 


and by transposing p, we shall have 


e=—p++/o+p’, and’ s= —p— 9+ p?. 
Either of these values being substituted for x in the equation 
a” -+- 2px = 9 
will satisfy it. For, we have from the first value, 
@=(—ptVgt+py= P—»Pv¢t+Pt+ot+P 

and | She ee 

2px = 2p x (—p+ Vat Pp) =—wt+pVvat+P 
hence a +. 2px = q. 


136 ELEMENTS OF ALGEBRA. (CHAP. v1, 


For the second value, we have 


®@—=(—p—VqtpPyP= p+%»Vgt+pPt+qt+P 


and 
2px = 2p(—p—Vq+p’) =—w—wpV/g +p; 
hence, ie” pie sages 


and consequently, the values found above, are roots of the equa- 
tion. | 

In order to refer readily, to either of these values, we shall call 
the one which arises from using the + sign before the radical, 
the first value of x, or the first root of the equation; and the 
other, the second value of a, or the second root of the equation. 

Having reduced a complete equation of the second degree w 
the form 

2a ¢, 

we can write immediately the two values of the unknown quan- 
tity by the following 


RULE. 


The first value of the unknown quantity is equal to half the co- 
efficient of « taken with a contrary sign, plus the square root of 
the absolute term increased by the square of half the co-efficient of a. 

The second value of the unknown quantity is equal to half the 
co-efficient of « taken with a contrary sign, minus the square rool 
af the absolute term increased by the square ‘of half the co-efficient 
OF a 

1. Let us take as an example the equation 

x? — Te +10=—0. 

By transposing 10, we have 


a? — Tx = — 10. 
Hence, «2 = 3.5 + +/— 10+ (3.5)? — 3.5 + +/2.25 =5, 
and w= 35 — 4/— 10-1 (3.5) = 35 oa 
2. As a second example, let us take the equation 
5 1 3 2 273 
th Pile Bhal wa 
5° 5 + r 8 3 Oe i 2" 


Reducing to entire terms, we have 
10x? — 6x + 9 = 96 — 8x — 12x? + 273, 










‘HAP. VI.] EQUATIONS OF THE SECOND DEGREE. 137 


nd transposing and reducing, 
| 22x? + 2x = 360, 
‘nd dividing both members by 22, 


2 360 
14 5 Naif OOM da 
Bea 5a agrs 
1 360... 71,2 
h - oy + (55) 
vines d 92 * V a9 + Nag 
1 360. %1\2 
and ee Se A (—) ; 
22 a2 * \aa 


» It often occurs, in the solution of equations, that p? and g are 
ractions, as in the above example. ‘These fractions most gen- 
rally arise from dividing by the co-efficient of #? in the reduc- 
jion of the equation to the required form. When this is the case, 
iwe readily discover the quantity by which it is necessary to mul- 
|iply the terms of qg, in order to reduce it to the same denominator 
‘with p?; after which, the numerators may be added together and 
placed over the common denominator. After this operation, the 





nt Sen 


‘lenominator will be a perfect square, and may be brought from 
}mder the radical sign, and will become a divisor of the square 
oot of the numerator. 

' To apply these principles in reducing the radical part of the 
hralues of a, in the last example, we have 


cbeeeer moo x eg 1* | (7920 Pa 
goue by w (22)? x (22)? (22)? 


1 89 
= 5 VM21= 53 
and therefore, the two values of x become, 
ee 08 Le 
Sera N82 92. se! 
and DRA D 
22 2 22 ih hg 


sither of which values being substituted for # in the given equa- 
ion, will satisfy it. 
3. What are the values of «2 in the equation 


ax?— ac = cx — ba?. 


138 ELEMENTS OF ALGEBRA. (CHAP. VI. 


We have, by transposing and reducing, » 


(a + 6) x? — ce = ac; 





hence x ne 
4 a bwin ee 
and consequently, 
2 
c ac c 
“= + ——_ —— +} 


2(a + 5) a+b' 4(a+b) 


an ar c ac c2 
7 =) ott). Ya 


If now, we multiply both terms of g by 4(a-+ 8), it will be 
reduced to a common denominator with p?, and we shall have 


ae ‘a c? ax; PS + 4abe +c? ye o/ 4a2e + 4abe+ a 
atb' 4(a+ 6? — 4 (ab) Taal by BERT 


¢ -& »/4a%e + 4abe + c2 
UE eS ae < 

















hence, 


2 (a+ b) 
| 4, What are the values of a, in the equations, 
6x? — 37x = — 57. 
By reducing to the required form, we have 
ibe ae 
lathe Or 


aie e DT a A Ye 
hence, x= = +V—=+ (5) 
12 6 12 


We observe, that if we multiply both terms of g by 2, and 
then by 12, that g and p? will have the same denominator; hence, 


_ 114 % 12. ee 
+ le, 114 x 12, (377 
12 (12)? (12) 
But, 114 x 12 = 1368; and (37)? = 1369; 


hence, z= aN 1808 360s 
| i2~ V (12) h42 12 


B7 (iileadge \y (ie 
19 ae ea ee 








HAP. VI.] EQUATIONS OF THE SECOND DEGREE. _ 139 


5. What are the values of a in the equation 


Aa? — 2x” + 2ax = 18ab — 1862. 


In this equation, the term which contains the second power of 

ie unknown quantity is negative; and since that term already 
tands in the first member of the equation, it can only be ren- 
ered positive by changing the sign of every term of the equa- 
on. Doing this, transposing, and dividing by 2. we have 


x* — ax = 2a? — 9ab + 96?; 


g 


2 


a 9a? 
aia, 4}. pe) de 2. 
a Sab + 9b?; 


2 
whence, 2#=—+ / 2a? — 9ab + 9b? + > 





2 Sas am t 





nd the root of the radical part is equal to _— 3b. Hence, 
a 3a x= 2a — 3b. 
te Cg 88) hence, ae Cee ae 
EXAMPLES. 
: x 4 
1. Given, ae 4 — x* — 2x — no = 45 — 32”, to find 2. 
e= 7.12 at 
| Ans. se od aaa to within 0.01. 
2. Given, «2? —8xr+10=19, to find «. 
Ans. x= 9, and e=—l1. 


. 3. Given, 2? — x — 40 = 170, to find a. 
Ans. x =15, and «= — 14. 


4, Given, 327+ 22 —9= 776, to find «x. 
Ans. x= 5, and «= — 52. 


5. Given, 427 —}3e+ 73 = 8, to find x. 


or 


| Ans. 2 = 14, and f= — 
6. Given, ma? — 2mxz+/ n = nx? — mn, to find x. 
fae ea 
Ans. «= , = 
NS. 2 ie a vee x ahaa v1 wa 














ELEMENTS OF ALGEBRA. 


[CHAP. Vz, 
90 a7 
7 Given me —_ ud: — ricki == U5 to- find a. 
3 
Ans. «= 4 =——. 
x >» & 3 


‘ ac 
. 8. Given ax? — ———= cx — bx?, to find a. 
a+ 6 


ct V+ 4ac 
Ans. x= aE I AENER as 
2(a + d) 
2 2 Ras 2 2 
~ 9 Given aby? + =< r= 6a! ee b?x 








An Bae ee 2m 
ac be 


2 m2 
“10. Given a? + 6? —2b4 4 2? = pes to find 2a. 
n i 


n 


Ans. 2 = (bn + if am? + bm? — a?n?). 
n*?— m? 





QUESTIONS. 


1. To find a number such, that twice its square added to thre 
times the number, shall be equal to 65. | 


Let x represent the number. Then, 


Qu? + 3x = 65, 


2) 65 9 3 lees 
= — — Ht — — = —— +t — 3 
whence, x T= 5 er Z rh 
therefore, 
a eae | 13 
sh otis neb ey iy — Fa and oe on: ry 


Both these values will satisfy the question, understood in it 
algebraic sense. We have, 


2x (5? + 3k OS 2 «25 4 aes 


13\2 13. 169 “SO Meare 
Suppose we had stated the question thus:—To find a numbet 


such, that three times the number subtracted from twice its square 
shall give a product equal to 65. 


‘VAP. VI.) EQUATIONS OF THE SECOND DEGREE. 14] 


‘If we denote the number by x we shall have 


2x2 — 3x = 65; 
es (6 Diu ee 428 
7 Peay fe Se 
hence, 4 a7] A Bd 
oA ot 23 
_erefore, e+ = 5° and cals Rap ye ere 


alues which differ from those found before, only in their signs. 
If the last enunciation be understood in its algebraic sense, the 


“5 equally with the + = will satisfy both the enunciation 


ad the equation. It is true that the second term — 3a will 
? added to the first term; for, the subtraction of 3 times — 5, 
all give + 15. | 


2. A person purchased a number of yards of cloth for 240 
ents. If he had received 3 yards less, for the same sum, it 
ould have cost him 4 cents more per yard. How many yards 
id he purchase ? 


Let «= the number of yards purchased. 


' Then, the price per yard will be expressed by ee 


Tf, for 240 cents, he had received 3 yards less, that is, « — 3 
ards, the price per yard, under this hypothesis, would have been 


240 | 
ypresented by a’ But, by the enunciation this last cost 


ould exceed the first, by 4 cents. Therefore, we have the equa- 
on 


hence, by reducing, «? — 32= ar APO: 


j 
ad eat 4180 =2 atl 





ierefore, w= 15, v x= — 12. 

|The value «—15 satisfies the enunciation understood in its 
' 240 

‘ithmetical sense; for, 15 yards for 240 cents, gives ae? Ane 

i . ; 





5 cents for the price of one yard, and 12 yards for 240 cents, 
ves 20 cents for the price of one yard, which exceeds 16 by 4 





ey eS ce 


142 ELEMENTS OF ALGEBRA. [CHAP. V1, 


The — 12 will satisfy the question in its algebrai¢ sense, and 
considered without reference to its sign, will be the answer to 
the following arithmetical question :—A person purchased a number 
of yards of cloth for 240 eents: if he had paid the same sum for 3 
yards more, it would have cost him 4 cents less per yard. How 
many yards did he purchase ? 


Remark.—In the solution of a problem, both roots of the equa- 
tion will satisfy the enunciation, understood in its algebraic sense. 

If the enunciation, considered arithmetically, admits of a double 
interpretation, when translated into the language of Algebra, the 
solution of the equation will make known the fact: and hence, 
while one root resolves the question in its arithmetical sense, the 
other resolves another similar question also in its arithmetical 
sense ; and both questions will be stated by equations of the same 
general form, having equal numerical roots with contrary signs. 


3. A man bought a horse, which he sold for 24 dollars. At 
the sale, he lost as much per cent. on the price of his purchase, 
as the horse cost him. What did he pay for the horse ? | 

Let « denote the number of dollars that he paid for the horse: 
then, x — 24 will express the loss he sustained. But as he lost a 


per cent. by the sale, he must have lost 7 upon each dollar, and 


2 
upon « dollars he loses a sum denoted by at we have then 


the equation 
2 


a 
TE ER Dd hha ee eS ; 
“Ti7) Vannes 24, waentS x 100x2 = — 2400; 
and x = 50 + VY — 2400+ 2500 = 50 + 10. 
Therefore, , «©=60 and «= 40. 


Both of these values satisfy the question. 

For, in the first place, suppose the man gave 60 dollars for 
the horse and sold him for 24, he then loses 36 dollars. But 
from the enunciation he should lose 60 per cent. of 60, that is, 
nitid of 60 Wo ftetse hee 36; therefore 60 satisfies the enu 
100 oc | 
ciation. Ls : 

If he pays 40 dollars for the horse, he loses 16 by the sale; 


® 


AP. VI.] EQUATIONS OF THE SECOND DEGREE. . 148 








ir, he should lose 40 per cent. of 40, or 40 x de —16; there- 


re 40 verifies the enunciation. 


| 4. A grazier bought as many sheep as cost him £60, and after 
“serving 15 out of the number, he sold the remainder for £54, 
id gained 2s. a head on those he sold: how many did he buy? 
Ans. 75. 
5. A merchant bought cloth for which he paid £33 15s., which 
je sold again at £2 8s. per piece, and gained by the bargain 
; much as one piece cost him: how many pieces did he buy? 
: Ans. 15. 
i 6. What number is that, which, being divided by the product 
* its digits, the quotient is 3; and if 18 be added to it, the 
igits will be inverted ? Ans. 24. 


7. To find a number such that if you subtract it from 10, and 
i the remainder by the number itself, the product shall be 21. 
Ans. 7 or 3. 
| 8. Two persons, A and B, departed from different places at the 
me time, and travelled toward each other. On meeting, it ap- 
eared that A had travelled 18 miles more than B; and that A 
ould have gone B’s journey in 15? days, but B would have been 
'g days in performing A’s journey. How far did each travel ? 
| Wie a 72 miles. 
| B 54 miles. 
9. A company at a tavern had £8 15s. to pay for their reck- 
ning ; but before the bill was settled, two of them left the room, 
nd then those who remained had 10s. apiece more to pay than 
‘efore: how many were there in the company ? Ans. 7. 


_ 10. What two numbers are those whose difference is 15, and 
f which the cube of the lesser is equal to half their product ? 
Ans. 3 and 18. 


, 11. Two partners, A and B, gained $140 in trade: A’s money — 
ia 3 months in trade, and his gain was $60 less than his stock ; — 
Ys money was $50 more than A’s, and was in trade 5 months: 
th hat was A’s kre Ans. $100. 
ha 


ey 


22. Two persons, A and B, start from two different points and 
‘avel toward each other. When they meet, it appears that A has 
‘avelled 30 miles more than B. It also appears that it will take A 


\ 





144 ELEMENTS OF ALGEBRA. _ (CHAP. VI 


4 days to travel the road that B had come, and B 9 days to trave 
the road that A had come. What was their distance apart wher 
they set out ? Ans. 150 miles. 


Discussion of Equations of the Second Degree. 


141. Thus far, we have only resolved particular problems in 
volving equations of the second degree, and in which the knowr 
quantities were expressed by particular numbers. 

We propose now, to explain the general properties of thes¢ 
equations, and to examine the results which flow from all the sup- 
positions that may be made_on the values and signs of the knowr 
quantities which enter into them. 


142. It has been shown (Art. 140), that every complete equa: 
tion of the second degree can be reduced to the form 


a? +- 2pe = 9 a4 

in which p and gq are numerical or algebraic quantities, whole 
numbers, or fractions, and their signs plus or minus. 

If we make the first member a perfect square, by adding p? te 
both members, we have 

oP ape tp? ae 
which may be put under the form 
(2+ pp)? =a p*. 


Whatever may be the value of g + p%, its square root may be 
represented by m, and the equation put’ under the form 


(a + p)? = m?, and consequently, (# + p)? — m?=—0. : 


But, as the first member of the last equation is the differences 
between two squares, it may be put under the form 


(x + p—m)(x+ p+m)=0, (2) 

in which the first member is the product of two factors, and the 
second 0. Now we can make the product equal to 0, and cor 
sequently satisfy equation (2), only in two different ways: viz. 
making 

x-+p—m=—0O, whence, «= —p+am, | 
or, by making 

e-+ip-+m = 0, whence, # = —p—m; 


“HAP. VI.] EQUATIONS OF THE SECOND DEGREE. 145 


id by substituting for m its value, we have 


z= —p+/q+p?, and x= —p—Vo+p?. 

Now, either of these values being substituted for 2 in its cor-, 
‘sponding factor of equation (2), will satisfy that equation, and 
pnsequently, will satisfy equation (1), from which it was” derived. 
ence we conclude, 





t. That every equation of the second degree has two roots, and 
‘i two. 

(2d. That every equation of the second degree may be decomposed 
to two binomial factors of the first degree with respect to x, having 
for a first term, and the two roots, taken with their signs changed, 
or the second terms. 





For example, the equation 

x? + 32 — 28= 0 
>ing resolved gives x = 4 and « = —7; either of which values 
ill satisfy the equation. We also have 
) (a — 4) (a+ 7) = 2? + 3a — 28. 
If the roots of an equation are known, we readily form the bi- 
mnial factors and the equation. 


1. What are the factors, and what is the equation, of which. 
e roots are 8 and — 9? 


x«—8 and «+9 
‘e the binomial factors, and 

(a9 — 8)(#+9)=2?+2—72=0 
the equation. 


2. What are the factors, and what is the equation, of which 
e roots are —1 and +1. 


(2+ 1)(rx—1)=2?—1=0. 
3. What are the factors and what is the equation, whose roots 
e 





Pome 1030... H-4/.—1039 

ee IG, uw 

TpH/ D089 7 = 4f'= 1039 
(»— 16 —) x (2- 16 =) 


= 82? — 7x + 34—0. 
10 


146 ELEMENTS OF ALGEBRA. [CHAP. Vi 


143. If we designate the two roots of any equation by 2’ an 
, we shall have 


va p+ Vath and =p Vet 
by adding the roots, we obtain, 

a’ + a” = — 2p; 
and by multiplying them together, 


ae’ / 


a’a’’ = — gq. Hence, 


Ist. The algebraic sum of the two roots 1s equal to the co-efficien 
of the second term of the equation, taken with a contrary sign 

2d. The product of the two roots is equal to the absolute term 
taken also with a contrary sign. 


144. Thus far, we have regarded p and g as algebraic quanti 
ties, without considering the essential sign of either, nor have wi 
at all regarded their relative values 

If we first suppose p and gq to be both essentially positive, the 
to become negative in succession, and after that, both to becom 
negative together, we shall have all the combinations of sign 
which can arise; and the complete equation of the second de 
gree will, therefore, always be expressed under one of the fou 
following forms :— 


ah ope — 


ef —2or =) 9 (2), 
x? + 2px = — gq (33 
a? — 2px = —gq (4). 


These equations being resolved, give 


a= pty g+pt 1), 
e= tp omen 
e=—ptv—g+p (3), 
w= +p+/—qt+p? (A). 
In order that the value of x, in these equations, may be foun/ 


either exactly or approximatively, it is necessary that the quam 
tity under the radical sign be positive (Art. 126). 


Now, p® being necessarily positive, whatever may be the sigi 
of p, it follows, that in the first and second forms all the value: 













JAP. VI.] EQUATIONS OF THE SECOND DEGRER. LAy 


| @ will be real. They will be determined exactly, when the 
|1antity q+ p® is a perfect square, and approximatively, when it 
not so. 


| dical expression + of q+ p® will be greater than p, and hence 
e second member of the equation will have the same sign as 
e radical. ‘Therefore, in the first form, the first root of the equa- 
m will be positive, and the second root negative. 


The positive root will, in general, as already observed, alone 
tisfy the problem understood in its arithmetical sense ; the neg- 
ive value, answering to a similar problem, differing from the first 
lily in this; that a certain quantity which is regarded as addi- 
\re in the one, is subtractive in the other, and the reverse. 

i In the ‘second form, the first value of 2 is positive, and the 
jcond negative, the positive value being the greater. 

‘In the third and fourth forms, the values of x will be imaginary 
then g > p?, and real when quik pre 


‘And since af —q+p*? <p, it follows that the real values of 





/ 


| 145. The last properties which have been proved, may be shown 
| om the two properties of an equation of the second degree, de- 
aed in Art. 143. They are: 


| The algebraic sum of the roots is equal to the co-efficient of the 
jcond term, taken with a contrary sign, and their product is equal 
the absolute term, taken also with a contrary sign. 


| In the first’ two forms, q being positive in the second member, 
| follows that the product of the two roots is negative: hence, 
\2y will have contrary signs. But in the third and fourth forms, 
being negative in the second member, it follows that the prod- 
|t of the two roots will be positive: hence, they will have like 
ms, viz., both negatwe in the third form, where 2p is positive, 
| d both positive in the fourth form, where 2p is negative. 
|, Moreover, since the sum of the roots is affected with a sign 
| ntrary to that of the co-efficient 2p, it follows that the nega- 


,e root will be the greatest in the first form, and the least in the 
jrond. . 


148 ELEMENTS OF ALGEBRA, (CHAP. VI 


146. We will now show, that when in the third and fourt} 
forms the roots become imaginary, that is, when g > p?, that thi 
conditions of the question will be incompatible with each other 
and therefore, the values of x ought to be imaginary. — 

Before showing this, it will be necessary to establish a propo 
sition on which it depends: viz., 

If a given number be decomposed into two parts, and those part: 
multiplied together, the product will be the greatest possible when th 
parts are equal. 

Let a be the number to be decomposed, and d the difference 
of the parts. ‘Then 


= + = = the greater part (Art. 32). 
a d 
and ey Mies ak gs the less part. 
2 
and > — Ee =P, their product (Art. 46). 


Now it is plain, that P will increase as d diminishes, and thai 
it will be the greatest possible when d= 0; that is, 
a a a® 


— X — = — is the greatest product. 
eae Sir ae) gt P 


147. Now, since in the equation 


x? — 2nn = —q 


2p is the sum of the roots, and g their product, it follows that , 
cannot be greater than p?. ‘The relations between p and gq, there- 
fore, fix a limit to the value of q; and if we assume, arbitrarily 
gq > p*, we express by the equation a condition which cannot be 
fulfilled, and this contradiction is made apparent by the values of 
x becoming imaginary. Analogy would lead us to conclude that 


When the value of the unknown quantity is found to be imagir 
ary, the conditions expressed by the equation are incompatible with 
each other. 


Remarx.—Since the roots, in the first and second forms, have 
contrary signs, the condition that their sum shall be equal to a 
given number 2p, does not fix a limit to their product: hence, im 
those two forms the roots are never imaginary. 


jmp. VI.) EQUATIONS OF THE SECOND DEGREE. 149 


, 148. We shall conclude this discussion by the following re 
harks :— 

Ist. If, in the third and fourth forms, we suppose g = p?, the 
jidical part of the two values of x becomes 0, and both the values 
lsduce to x= = p: the two roots are then said to be equal. 


t In fact, by substituting p* for g in the equation, it becomes 
a? + 2px = — p®, whence 

| a? + 2pa + p? = 0, that is, (4 + p)? = 0. 

| Under this supposition, the first member becomes the product 

\f two equal factors. Hence, the roots of the equation are equal, 


|ince the two factors being placed equal to zero, give the same 
Jalue for 2x. 





2d. If, in the general equation, 
il x“? + 2p2 = gq, 
re suppose g = 0, the two values of x reduce to 
I, 2=—p+p=—0, and x= —p— p= — 2p. 
' Indeed, the equation is then of the form 
i x2? + 2px = 0, or x(x + 2p)=—0, 
rhich can only be satisfied, either by making 

ew 0; or # +270); 


| whence, x2=0, and «= — 2p; 





liat is, one of the roots is 0, and the other the co-efficient of x, 
jUken with a contrary sign. 


3d. If, in the general equation 





: Roe apy ox ot 9, 
\'e suppose 2p = 0, there will result 


masta g/ owhence,: a# ic +f 9,5 





jiat is, in this case the two values of x are equal, and have con- 
lary signs, real in the first and second forms, and imaginary in 
lie third and fourth. 

The equation then belongs to the class of equations involving 
|vo terms, treated of in Art. 139. 

| 4th. Suppose we have at the same time p= 0, g= 0; the equa- 
on reduces to 2? = 0, and gives two values of x, equal to 0. 


150 ELEMENTS OF ALGEBRA. [CHIAP. ¥1 


149. There remains a singular case to be examined, which i 
often met with in the resolution of problems involving equation 
of the second degree. 

To discuss it, take the equation 
ax* + br = ¢, 

— b+ Jb + 4ac 
ee 

Suppose, now, that from a particular hypothesis made upon th 
given quantities of the question, we have a= 0; the expressio 


which gives eae 


for «x becomes 


—— ‘0 
e= — shad whence "ae 
= —— c 
Vins ‘ 26 
Ff Sipe ar 0” 
Let us first interpret the first root of = 3° 


By multiplying the numerator and denominator of the secor 
member of the equation 


ed h2 
Ris WEBI Cy Pose by «1 bs a ef ens 














2a 
we obtain 
b2 — (b? + 4ac) — 4ac 
Ble —— SS 
2a (— b —4/b?+-4ac) 2a (— b — Vb? + 4ac) 
— 2c 
hence, een =e) Oe 
—b—v0h+ 4ac af nee 
and consequently, Boos =, by making a= 0. 


Hence we see that the apparent indetermination arises from 
common factor in the numerator and denominator. 
In regard to the second root 


we see that it is presented under the form of infinity. By maki 
a = 0, in the equation 

ax? + bx =e, 
it reduces to an equation of the first degree, ; 


bx = c. 


if 
| 

| 
i 

i 


UHAP. VI.) |= EQUATIONS OF THE SECOND DEGREE. 151] 


_ It is therefore wnpossible that it can have two roots; and hence. 
uch a supposition gives one of the values of & infinite. 

| We have already seen (Art. 147), that wnaginary values of the 
nknown quantity indicate the introduction, into the equation, of 
ontradictory conditions. By considering the above discussion, and 
aat of Art. 110, we would conclude, that a result which is in- 
nite, indicates the introduction into the equation of a condition that 
s absolutely impossible. 

' If; we had at the same time 


j : mice: (), b-=.0, CoO), 


ne proposed equation would be altogether indeterminate. ‘This is 
ae only case of indetermination that the equation of the second 
legree presents. 

We are now going to apply the principles of this general dis- 
‘ussion to a problem which will give rise to most of the circum- 
tances that are commonly met with in problems involving equa- 
ions of the second degree. 


: 
Lig 


Problem of the Lights. 


See ae CRB, CF 

150. Find upon the line which joins two lights, A and B, of 
ifferent intensities, the point which is equally illuminated; ad- 
litting the following principle of physics, viz: The intensity of 
ihe same light at two different distances, is in the inverse ratio of 
he squares of these distances. 

Let the distance AB, between the two lights, be expressed by 
; the intensity of the light A, at the units distance, by a; that 
if the light B, at the same distance, by 4. Suppose C tobe the 
qually-illuminated point, and make AC = «, whence BO =c — 2a. 

By the principle we have assumed, the intensity of A, at the 
nity of distance, being a, its intensity at the distances 2, 3, 4, 
fee. will be 7, ee 

j 


‘ ; a ted : 
‘@ expressed by —-. In like manner, the intensity of B at the 
p y oes : ) y 


&c.; hence, at the distance z it will 


a ; b a; 
sistance c — x, is (6 ay but, by the conditions, these two 
> +. c— x 





152 ELEMENTS OF ALGEBRA. (CHAP. Vu, 


eee aan 
O4 A omar a 
intensities are equal to each other, and therefore we have the 
equation 
a b 
a? (¢ — @)?’ 
which. can be put under the form 
(Cueto 
ie lpn 


pag i +b whence, 
Fionoie ie 
s Way which gives, ee iy ae 
aya 
Tas | 


b 
a 








Hence, 


Ist root 1s, «= 











2d root is, #»—=—=——, which gives, c—# = 
b 


Ist. Suppose a> Ob. 


The first value of 2 is positive; and since 


Va 
etree oy af 
Va+tyv b 
it will be less than ¢, and consequently, the required point C 


will be situated between the points A and B. We see moreovei 
that the point will be nearer B than A; for, since a>b, we 


have 
vatva or, 24/4 > \1/.aeee J b), whence, | 


Page oe a cya c 
> 5 and consequently, ——=————= > —. 
Bc Og I Vatyb” 2 
Indeed, this ought to be the case, since the intensity of A 
supposed greater than that of B. 


The corresponding value of c—a, as may be easily shown, is 
also positive, and less than one half of c; that is, 


cb c 
Japs ae 


The second value of « is also positive; but since, — 












i’ 
‘HAP. VI.) EQUATIONS OF: THE SECOND DEGREE. 153 


‘al 
ge 


will be greater than c; and consequently, the required point 
ill be at some point C’, on the prolongation of AB, and at the 
ght of the two lights. 

We may, in fact, conceive that since the two lights exert their 
luminating power in every direction, there should be upon the © 
rolongation of AB, another point equally illuminated; but this 
oint must be nearest that light whose intensity is the least. 

We can easily explain, why these two values are connected 
y the same equation. If, instead of taking AC for the unknown 
uantity a2, we had taken AC’, there would have resulted 
ik =a«—c; and the equation 


a b 


a (@—e)? 


=~ 


Tow, as ( — c)? is identical with (c — «)*, the new equation is 
lentical with that already established, which consequently should 
ave given AC’ as well as AC. 
And since every equation is but the algebraic enunciation of a 
roblem, it follows that, when the same equation enunciates several 
roblems, wt ought by tts different roots to solve them all. 
When the line AC’ is represented by the unknown quantity 2 
oth members of the equation 

EB ciy/ bh 


He eS == SS ee 
ay a—v b 
re negatived, as they ought to be, since x >. 
| By changing the signs of both members, we have 


c— 





cv b 
— a —$<$—<—_— —= ABCt 
x Cc amy e7 j 
2d. Let a<b. 


| This supposition gives a positive value for 


Ind since Jat+VJ/b>/a4+7a, that is," >2Jfa 





: 


154 ELEMENTS OF ALGEBRA. (CHAP. VI, 


a a 
C” A C B C’ 
it follows that, acs, 
c 
and consequently, Ft 3°? 


and therefore, under this hypothesis, the point C, situated between 
A and B, will be nearer to A than B, as indeed it ought, since 
the feebler light is at A. 

The second value of a, that is, 


ASA eine cla 
Joni. 
is essentially negative. How is it to be interpreted ? 

Let us suppose that we had considered C”, at the left of “A, 
as the point of equal illumination, and that we had represented 
AC” by —2. 

Then, - BC” = BA + ACG®™ 

that is, BC’ =c+(—2)=¢ = 2} 





and the equation of the problem would be 
a b a b 


a sa BE that is, = (oe 

and therefore, this equation ought to give the point C” which lies 
to the left of A, as well as the points C and C’ which lie to the 
right. 

It should be observed, that we have regarded — a, which rep- 
resents AC”, as a mere symbol, without reference to the essential 
sign of x. Indeed, the essential sign of the ‘unknown quantity 1s, 
in general, only made known in the final result. 

If it appears, in the final result, that a #tself is negative, the 
numerical value of, : 


BC” =c+(—«x) becomes BC’ =c+2; 
that is, BC” will be equal to ¢ plus the numerical value of i 
or to ¢ minus its algebraic value. Hence, : 


Bo? = ae a i | 





a quantity which is essentially positive. 


a tell 
y 
' 


‘HAP. VI.] EQUATIONS OF THE SECOND DEGREE. 155 


i. 3d. Let == b. 

_ Under this supposition, the value of a, and that of c —2, for 
| 

he point C between A and B, both reduce to = that is, when 
he lights are of equal intensity, the point of equal illumination is 
at the middle of the line AB. 

igi he value of x, and that of c—a, for the points C’ and ©”, 
which lie on the prolongation of AB, both reduce to 


( A ee —eVb 


a see 
Ta or, to 6 , that is, to infinity ; 








which indicates, that the conditions of the question are absolutely 
impossible. It is evident, indeed, that they are so, for, when the 
intensity of the two lights is equal, no part lying on the prolon- 
ation of AB could be as much illuminated by the distant as by 
the nearer light: hence, the supposition of equal illumination, from 
which the equation of the problem is derived, is dnpossible; and 
this is shown in the analysis by the corresponding values of the 
unknown quantity becoming infinite. 


4th. Let a=b, and c= 0, 


_ Under these suppositions, the value of # and of ¢ — a, for the 
point of equal illumination between A and B, both reduce to 0, 
as indeed they ought to do, since the points A, B, and C, are 
then united in one. 

. The value of x, and of c—a, for the points C’ and C”, re- 
duce to the indeterminate form 


0 
Wey 
' Resuming the equation of the problem 
| (a — b) a? — 2acr = — ca, 


we see that it becomes, under the above suppositions, 
0.22 — 0.2 = 0, 


Wich may be satisfied by giving to x any value whatever: hence, 
it is a case of indetermination. Indeed, since the two lights are 
of the same intensity, and are placed at the same point, they ought 
la illuminate equally every point of the straight line. 





J f 


\ 





156 ELEMENTS OF ALGEBRA. (CHAP. VI. 


5th. Let c= 0, and a and b be unequal. 

Under this supposition, both values of x, and both values of 
c — x, will reduce to 0; and hence, there is but one point of the 
line that will be equally illuminated, and that is the point at which 
the two lights are placed. | 

In this case, the equation ‘of the problem reduces to 


(a —b) x? = 0, 
which gives two values, 
x=0, and «=0. 
The preceding discussion presents a striking example of the 
precision with which the algebraic analysis responds to all the 


relations which exist between the quantities that enter into the 
enunciation of a problem. 


Examples involving Radicals aa the Second Degree. 





1. Given, c+ fV@?+2? = Ts = to find a. 
a 
By reducing to entire terms, we ao 
aVva+t a+ a?+ x? =,ee", 

by transposing, 2 Ja? + x = a? — 2%, _ 
and by squaring, ara? + xt = at— 2a*x? + at, 
hence, 3a*a =a", 
and consequently, Speen = < 








: Ve ; /# s 
2. Given, tk apa aia = 0b, to find z. 


at SO a? hs 
By transposing, Nicte un =! oe +4; 

; a? a az 
and by squaring, mete f= — $2 + 2h eter b2; 
hence, eo ae and b= 2/52; | 
| 
; 
| | 
: 


and by squaring, b2 — —_— 422; 





}HAP. VIL] EQUATIONS OF THE SECOND DEGREE. 157 
2 2 
a 
nd hence, x =—, and «=+ a 
5b? by 5 
pag 

















“$. Given, =——-—— = §, to find xz. 





























Ans. «= et ial 
1+) 
- 6. Given Vet Vo—a__ nto to find 
, & © : = , to find x 
: Vv —vyr-—a “2—a 
2 
i Ans eat) 
My 1+2n 
’ 7. Given, ee ase to find x 
Vx Vv x b 


Ans. © = +2 Jab — 92. 


8. Given a+a+ Y2ax + 2 


=, to find z. 
at+@ 





Of Trinomial Equations. 
151. Every equation which can be reduced to the form 
me? -|- 2pn” == q, 
a which m and n are positive whole numbers, and 2p and gq, 


mown quantities, is called @ trinomial equation. 
| ee a trinomial equation contains three ae of terms : viz. 







158 ELEMENTS OF ALGEBRA. [CHAP. VI, 


If we suppose m= 2 and n=1, the equation becomes 


x? + 2ne Sigs 


a trinomial equation of the second degree. , 


™ ¢ 

152. The resolution of trinomial equations of the second degree, 
has already been explained, and the methods which were pursued 
are, with some slight modifications, applicable to all trmomial equa- 


tions in which m = 2n, that is, to all equations of the form y 
2" -|- Qe ae | 
Let us take, as an example, the trinomial equation of the fourth 
degree, 
cat — dz? —axt+f=7+6. j 
—d 7+6— 
We have, gt + —_— x? = 7 ae : - 
c—a c—a 


and by substituting 2p for the co-efficient of x?, and q for the ab- 
solute term, we have y 
a* + 2pu? = q. 
If now, we make 
a* = y, and consequently, «2 = = one y 
we shall have ‘f 


y? + 2py = q, and y=—peVotp: 


hence, w= t4/—ptV/qt+pr 


We see that the unknown quantity has four values, since each 
of the signs + and —, which affect the first radical can be com- 
bined in succession with each of the signs which affect the sec- 





ond; but these values taken two and two are numerically equal, and 
have contrary signs. 


EXAMPLES. 
1. Take the equation 
at —~ 2502 = —+.144.45 t 
If we make a? — y, the equation becomes, a 
y? — 25y — — 144, 
which gives, y =2.16,, and, yeas 


" 


i 
F 


HAP. VI] EQUATIONS OF THE SECOND DEGREE. 159 


. Substituting these values, in succession, for y in the equation 
2 = y, and there will result, 


i ist. 2? = 16, which gives r=+4 and «x= —4. 
[2d 2? = 9, which gives r—+3 and #= — 3. 
dence, the four values are + 4, —4, + 3, and — 3. 
. 2. As a second example, take the equation 

at — Tx? = 8. 
If we make x? = y, the equation becomes, 


y? — Ty = 8, 





ie gives wee 8; Pend y= I, 


_ Substituting these values, in succession, for y, and we have 


(st. a= 8, which gives w=4+2+/ 2, and r=—2y 2. 
2d. a= —1, which gives ex=+ ees and «= — Aye 


The last two values of x are imaginary. 





: 3. Let us take the literal equation - 
at — (2be + 4a”) a? = — Be?, 
By making 2? =y, we have 
y? — (2be + 4a?) y = — be?; 
whence, y = be + 2a? + 2a J be + a; 


ind consequently, 


Hiei be + 2a? + 2a +/be + a. 
4, Suppose we have, 
’ 22—T 7 x = 99. 
If we make Ja —y, we have «= y?, and hence, 


i: 
. 2y? — Ty = 99; 
yom which we obtain 
11 
— 9 d —_—- — — 
y ’ an y 2 
12} 
hence, @=—81, and «= "EN 


153. Before resolving the general case of trinomial equations, 
t may be well to remark that, the nth root of any quantity, is an 
|:wpression which muitiplied by itself n—1 times will produce the 
fywen quantity. 





. 
) 


160 ELEMENTS OF ALGEBRA. [CHAP. VI. 


The method of finding the nth root has not yet been explained 
but it is sufficient for our present purpose that we are able to in- 
dicate it. 

Let it be required to find the values of y in the equation 

Yt st A ee 

If we make y*= 2, we have y? = a, and hence, the giver 

equation becomes 


x +. 2pn = q, 
and hence, e=—ptvVqt+p’; 
that is, y® = —p+/qt+p, 


and yt) =e ae 
If we suppose n= 2, the given equation becomes a trinomia 
equation of the fourth degree, and we have 


(ga h\f—p+ vote 


154. The resolution of trinomial equations of the fourth degree 
therefore, gives rise to a new species of algebraic operation: viz. 
the extraction of the square root of a quantity of the form 


at Jb, 


in which a and 6b are numerical or algebraic quantities. 
To illustrate the transformations which may be effected in ex: 


pressions of this form, let us take the expression 3 + 4/ 5. 
By squaring it we have 


(B+/5P=94+6/5 +5=1446/5: 


hence, reciprocally, 4/14+6/5 =32/5. 


As a second example, we have 


(f7+J7llpP=7+2/f774+11 = 1842/77: 


hence, reciprocally, 4/18 +2 TT oe af OF a AY Gi 


Hence we see, that an expression of the form 


Vox vis 


may sometimes be reduced to the form 


gt /V or Va+J/t; 








} 
| 
1 


‘AAP. V1.) EQUATIONS OF THE SECOND DEGREE. 161 


ad when this transformation is possible, it is advantageous to 
Tect it, since in this case we have only to extract two simple 
yuare roots ; whereas, the expression 


\/azJ/b 


squires the extraction of théMequare root of the square root. 


_ 155. If we represent two indeterminate quantities by p and 4, 
‘e can always attribute to them such values as to satisfy the 


ptqg=\/at+ 7b - - - (1), 
i p—-q=\/a-VJb - - - (2). 


These equations being multiplied together, give 


P-P=Ve—b - - - - (3). 

Now, if p and g are irrational monomials involving only single 
wdicals of the second degree, or, if only one is irrational, it fol- 
‘ws that p? and gq? will be Patient’ in which case, p? — q?, or 
s value, +a? — J, is necessarily a rational quantity, and conse- 
iently, a? — 6 is a perfect square. 

Under this supposition, a transformation can always be effected 
sat will simplify the expression. 

By squaring equations (1) and (2), we have 

P+ egt+Paaty d 
pe —2qt+ePaa—v ob, 
id by adding member ito member, 
PP yg? = a, 4-1 (A). 
_If we denote the second member of equation (3) by c, we shall 
ive 
F aoe =) SS). 

; By adding the two last equations and subtracting equation (5) 

om (4); we have 


quations 





oe = @-4-¢, and 297 = a—'e. 
ate ay 
2 2 


i therefore, p=\/ —-—, and g = soe 
) 


: 


162 | ELEMENTS OF ALGEBRA. (CHAP, Vi 


and consequently, 


Ava+t vy 6, or ptq=tV it 











and Ri af) eed Ramo YE - - (7). 


These two formulas can be verified; for by squaring both mem 
bers of the first, it becomes 


ay | ge Sis 
at Joa tite aegis c = =atya—c; 


but, Va—b=c, gives c?=a*—b. 
Hence, a+ J/b=at+Ve—@+baaty b. 


The second formula can be verified in the same manner. 








Remarx.—156. Formulas (6) and (7) have been deduced with- 
out reference to any particular value of ¢; and hence, they are 
equally true whether ¢ be rational or irrational. If, however, c is 
irrational, they will not simplify the given expression, for each 
will contain a double radical. Therefore, in general, this trans- 
formation is not used, unless a? — 0 is a perfect square. 


EXAMPLES. 


1. Reduce 4/ 94 + 42 peiy | 94 + »/ 8820, to its sim- 


plest form. We have - a=94, 6 = 8820, 
whence, hee Ja —d- —b= 4/8836 — 8820 =— 4, 


a rational quantity ; therefore, formula (6) is applicable to this 
case, and we have a 


s/ site 5 Be es fet a 


or, reducing, = + (4/49 + / 45) ; 

















i 
HAP. V1.] EQUATIONS OF THE SECOND DEGREE. 163 


¢ 
t 


erefore, (+275 + 42 V5 = ET +3 V5 5 ). 


deed, a 
(7434/5 P= 494 45 442 (5 = 04 + 42/5), 


| 2. Reduce 4/ np+ a 7 2m ~np +m, to its simplest 


‘mm. We have 


a= np + 2m?, and b= 4m? (np + m), 





a?’ — b=n’"p?, and c= Va—b=np; 
‘nd therefore, formula (7) is applicable. It gives, 


gee te skaruaialcensed ey 
; P 2 2 
ad, reducing, + (np + m — m). 


adeed, (o/ np +m? — m)? = np + 2m? — 2m /np + m. 


8. Reduce to its simplest form, 


Nee oy lat A / 16 — 307 —1. 


By applying the formulas, we find 
ye ees 

ad NY Wee 1 = 5 — 84/1: 

ee,» 4/ See 4) 16 — s0/ — 1 10. 


This last example shows very clearly the utility of the general 
‘oblem; because it proves that imaginary expressions combined 
gether, may produce real, and even rational results. 

4. Reduce to its ‘ines form, 


\/ 28 +1073. Ans. 5+4/3. 


5. Reduce to its simplest form, 


Md 30% | Aged aha/ =. 


6) Reduce to its simplest form, 


/ be + 2b Vie — + A/ be -- 2b be — D?. 


Ans. +26 

















164 ELEMENTS OF ALGEBRA. [CHAP. VI. 


7. Reduce to its simplest form, 


4/ ab + 4c? — d? — 2 / 4abc? — abd?. 


Ans. / ab — v/ 4c? — 


Equations of the Second Dede involving two or more 


Unknown Quantities. 





157. An equation involving two or more unknown quantities, 1s 
said to be of the second degree, when the greatest sum of the ew- 
ponents of the unknown quantities, in any term, ts equal to 2. Thus, 

3a? — 4a + y? — ay —dy+6=0, Tay —4e+y=0, 
are equations of the second degree. 

Hence, every general equation of the second degree, involving 
two unknown quantities, may be reduced to the form 

ay? + bry + cx? + dy + fe +g =0, 
a, », c, &c., representing known quantities, either numerical or 
algebraic. 

Take the two equations 

ayyt+tbay+cx?+dy+fe+g =0, 
ay? + Way+ ca? + dy+ fete =0. 

Arranging them with reference to x, they become 

ca?+(by+f)j)atay+dy+g =), 

fe + (byt fyetay+tdy+ey =9; 
from which we may eliminate «a?, after having made its co-efh 
cient the same in both equations. 

By multiplying the first equation by ej and the second by ¢, 
they become 

e/a? ++ (by + f)cex+ (ay +dy+g)c¢=0, 
c/a? + (Wy + fjcext (acy? +dyt+e’)e =0. 


Subtracting one from the other, we have 


[be — eb) y + fe — of at (ac — ea!) y+ (del — ei) y + 
— cy = 0, 





} 


which gives 


(ca’ — ac’) y* + (ed’ — de’) y + cy’ — ge’ 


go A 


(be? — cb’) y + fe — of’ 









I: 


HAP. VI.) EQUATIONS OF THE SECOND DEGREE. 165 


_ This value, being substituted for x in one of the proposed equa- 
ions, will give a final equation, involving y. 

_ But without effecting the substitution, which would lead to a 
very complicated result, it is easy to perceive that the final equa- 
ion involving y will be of the fourth degree. For, the numerator 
of the value of x being offhe form my? + ny + p, its square 
vill be of the fourth degree, and this square forms one of the 
yarts of the result of the substitution. 

Therefore, in general, the resolution of two equations of the sec- 
md degree, involving two unknown quantities, depends upon that of 
m equation of the fourth degree, involving one unknown quantity. 


158. The manner of resolving a general equation of the fourth 
legree, not having been yet explained, we cannot here give a 
complete theory of this subject. We will, however, indicate some 
of the particular methods by which equations of the second de- 
jree involving two or more unknown quantities, may be resolved 
vy an equation of the second degree involving but one. 

1. Find two numbers such, that the sum of the respective prod- 
icts of the first multiplied by a, and the second multiplied by 6, 
thall be equal to 2s; and the product of the one by the other 
qual to p. 


Let « and y denote the required numbers, and we have 











az + by = 2s, 
and, TY == p. 
From the first 
2s — ax 
i ay ; 
vhence, by substituting in the second, and reducing, 
ax® — 2sx = — bp. 
1 
“herefore, f= - me or o/ s? — abp, 


| onsequently, y= Gy A s? — abp, 


This problem is susceptible of two direct solutions, because 





ie Ve a abp, 


166 ELEMENTS OF ALGEBRA. [CHAP. VI 


but in order that the roots may ¥° real, it is necessary that 
s° > OL =aaop. 
Let a=b=1; the values of 2, and y, then reduce to 


wt ig a/'sth 7} and y=sy=vys*—p; 
whence we see, that under this s sition, the two values of ; 


are equal to those of y, taken in an inverse order; which shows 
that if 


s+ 4/s? —p represents the value of 2, s— + /s? —p j 


will represent the corresponding value of y, and reciprocally. 

This relation is explained by observing, that, under the las 

supposition, the given equations become 

ote y =) 2s and, gees ge 
and the question is then reduced to finding two numbers of whe 
the sum is 2s, and their product p, or in other words, to divide 
number 2s, into two such parts, that their product may be equal i 
a given number p. 

2. Find four numbers in proportion, knowing the sum 2s 0 
their extremes, the sum 2s’ of the means, and the sum 4c? 0 
their squares. 

Let u, x, y, z, denote the four terms of the proportion; th 
equations of the problem will be 


Ist condition, - - - u+z= 2s, 

2d condition, - - > e+ y = 2s’, 
since they are in proportion, Qe =r y, 
Ath condition, - - ue+ wo? + y? + 2? = 4c?. 


At first sight, it may appear difficult to find the values of th 
unknown quantities, but with the aid of an unknown auxihar 
they are easily determined. fs 

Let p be the unknown product of the extremes or means; 


shall then have 
zl etl Jar 
anal: which give, : av ged? 
us = D, a 
aie raed, which give, ; 7 Wag 3 P 
ey =p; y =s'— Vs? —p. 


and 


| 


‘. 
i 
i 
paar. VI.) EQUATIONS OF THE SECOND DEGREE. 167 


‘ Hence, we see that the determination of the four unknown quan- 
‘ities depends only upon that of the product p. 

| Now, by substituting these values of u, x, y, z, in the last of 
he equations of the problem, it becomes’, 


et V8 =p)? + (6- ere JF =p) 


4 

: $V ap Ae 

ind by developing and reducing, 

} As? 1. 45% — 4n — 4c?; hence, p = s?\- 5? — c?. 
Substituting this value for p, in the expressions for u, a, y, 2, 

‘we find 


bn Pe ea ems ++ ce — 8, 
= 3— 4/2 — s?, 


These four numbers evidently form a proportion; for we have 
uz=(s +e 2 9/2) (s— fe —s’ sg) = s*7 — c* + 8%, 
ay = (¢ + Ve — 5?) (x — Je —s?) =s?— 2 + 8? 


” Remarx.—This problem shows how much the introduction of 
‘an unknown auxiliary facilitates the determination of the principal 
unknown quantities. There are other problems of the same kind, 
‘which lead to equations of a degree superior to the second. and 
yet they may be resolved by the aid of equations of the first and 


‘second degrees, by introducing unknown auxiliaries. 


y= — / 2 — 82, 


3 Given the sum of two numbers equal to a, and the sum of 
their cubes equal to c, to find the numbers 


= 
| By the conditions a ee ee 


i 





ee + y3 =. 

Putting wc=s+z2, and y=s—z, we have a= 2s, 
x3 = 53 + 3522 + 352? + 29 

Yea. S\— Bs"z +) 382? — z: 


hence, by addition, 2° + y=2s3 + 6s22= 03 


and 

















168 ELEMENTS OF ALGEBRA. [CHAP, V1 


and by substituting for s its value, 


a c — 1a a Np 
oS —— ( )j=> te 
2 3a 2 12a 








4. Given, — ut Mes 48, and ee 24, to find x and Y- 


Ea HAS 











Dividing the first equation by the second, we have 


sedlbay panes and hence y = 4. 








and consequently, wWa=6, and «= 36. 


5. Given, « +VYay+y = 19 
a 6, ageiinates on to find w and y. 


Dividing the second equation by the first, we have 


e—Jryty= q; 





but, e+ fay +y=19: 
hence, 2x2 + 2y = 26 by addition, 
or, © hoe cena 
and _ 4/, xy + 13= 19 by substituting in the Ist eq.; 
or, /. ay 0 
and xy = 36. 
The 2d equation is, 2?-+ ay + y? = 133, 
and from the last, 3xy = 108; 





by subtracting x? — Qey + y? = 25: 





hence, C— y == Se. 
But, e+y= Ne Be 
hence, a2=9, or 4; and y=4, or 9. 


} 





‘ 
‘HAP. VI.] 


6. Find the values of x and y, in the equations 
a? + 32+ y = 73 — 2ay 
y? + 38y+ a= 44, 
By transposition, the first equation becomes, 
a? + 2ay ah 30 + y= 73; 
which, if the second be added, there results, 


w? + Qny + y? + 40 + 4y = (e+ y+ 4(e+y) = 117. 


If now, in the equation 


(x+y)? + 4(¢+ y) = 117, 


EQUATIONS OF THE SECOND DEGREE. 


169 


e regard «-+ y as a single unknown quantity, we shall have 


e-y= —2271174+4; 


‘hence, et+y=—24+11=9, 
and e-y= —2-—ll=—13; 
‘whence, e2=9—y, and x= —13—y. 


‘Substituting these values of « in the second equation, we have 


rier == oo, lor 2 = 9, 
and y? + 2y = 57, for «= — 13. 
The first equation gives, 


| y=5, and y= —7. 
id the second, 


y= — 1+ 4/58, and y= —1— +58. 


The corresponding values of x, are 


Wis AL fet Os 


j 


Ra Aaa ad x 1224/68. 


7 Find the values of x and y, in the equations 
xy? + xy? + xy — 600 — (y + 2) ay? 
ety=—14—y. 
From the first equation, we have 
: ay? +- (y? + 2y) x24? + xy? + xy = 600, 
, wy? (1 + y? + 2y) + xy (1 + y) = 600, 


| again, § xy?(1 + y)? + ay(1+ y) = 600; 


om 


170 ELEMENTS OF ALGEBRA. (CHAP. Vi} 


which is the form of an equation of the second degree, by re 
garding «y(1-+ y) as the unknown quantity. Hence, 
ae are /2401 
ey(1+y)=—}+ /600+}4=—44 eae? 
and if we discuss only the roots geen belong to the + valu 
of the radical, we have 





49 
and hence, ss all - 
ye 


Substituting this value of « in the second equation, we have — 


(~tyy—U4yY+y)=— a4; 


whence, yty=12, and ¥+-y= oe 
From the first equation, we have 
1 z 
Y er ia! ohne or —4; 
and the corresponding values of a, from the equation 
24 
e = ——— = 2 
yy 


From the second equation, we have 


ys 1, and) y= we 





which gives op al | 
8. Given, xy +ay?=6, and ay? + wy3 = 12, to find 


and y 
e=2 or 


ate. ca or 2. 


2 ae 
9. Given, ss ieee to find x and y. ! 
ey ==. 6 
ae or 2; of —=StwW/ BS 
Ans. 
y= 2, or 3; or —34Y¥ 3. 
QUESTIONS. 


1. There are two numbers whose difference is 15, and ha 


their product is equal to the cube of the lesser number. Wh 
Ans. 3 and 18. 


bal 


are the numbers ? 















‘CHAP. VI.1 EQUATIONS OF THF SECOND DFGREE. 171 


2. What two numbers are those whose sum multiplied by the 
greater, is equal to 77; and whose difference, multiplied by the 


‘lesser, is equal to 122 
Ans. 4 and 7, or 3+ 2 and } ¥ 2. 


3. To divide 100 into te such parts, that the sum of their 
‘square roots may be 14. | Ans. 64 and 36. 


_ 4. It is required to divide the number 24 into two such parts, 
‘that their product may be equal to 35 times their difference. 
Ans. 10 and 14. 


5. The sum of two numbers is 8, and the sum of their cubes 
jis 152. What are the numbers ? Ans. 3 and 5. 


| 6. The sum of two numbers is 7, and the sum of their 4th 
powers is 641. What are the numbers? Ans. 2 and 5. | 


| 7. The sum of two numbers is 6, and the sum of their 5th 
| powers is 1056. What are the numbers? Ans. 2 and 4. 


| 8. Two merchants each sold the same kind of stuff: the sec- 
jond sold 3 yards more of it than the first, and together, they re- 
ceived 35 dollars. The first said to the second, “I would have 
‘received 24 dollars for your stuff.” The other replied, “ And I 
| would have received 121 -doilars for yours.” How many yards 
/did each of them sell? - 

, tise si merchant agate - ig id 
. 7X: y= 18 y= 8. 

} 9. A widow possessed 13,000 doilars, which she divided into 


incomes from them were equal. If she had put out the first por- 
[tion at the same rate as the second, she would have drawn for 
| this part 360 dollars interest; and if she had placed the second 
{out at the same rate as the first, she would have drawn for it 
| 490 dollars interest. What were the two rates of interest ? 


Ans. 7 and 6 per cent. 


2 ELEMENTS OF ALGEBRA. CHAP. Vil. 


CHAPTER VII. 


OF PROPORTIONS AND PROGRESSIONS. 


159. Two quantities of the same kind may be compared to. 
gether in two ways :— 


Ist. By considering how much one is greater or less than the 
other, which is shown by their difference; and 


2d. By considering how many times one is greater or less than 
the other, which is shown by their quotient. 

Thus, in comparing the numbers 3 and 12 together with re- 
spect to their difference, we find that 12 exceeds 3, by 9; and in 
comparing them together with respect to their quotient, we find 
that 12 contains 3, four times, or that 12 is 4 times as great as 3. 

The first of these methods of comparison is called Arithmetical 
Proportion; and the second, Geometrical Proportion. Hence, 

ARITHMETICAL Proportion considers the relation of quantities 
to each other, with respect to their difference; and GEOMETRICAL 
Proportion, the relation of quantities to each other, with respect to 
their quotient. 


Of Arithmetical Proportion. ~ an | 


160. If we have four numbers, 
2, 4, 8, and 10, 


of which the difference between the first and’ second is equal to 
the difference between the third and fourth, these numbers are 
said to be in arithmetical proportion. The first term 2 is called 
an antecedent, and the second term 4, with which it is compared, 
a consequent. ‘The number 8 is also called an antecedent, and 
the number 10, with which it is compared, a consequent. The 


‘HAP. VII. ARITHMETICAL PROGRESSION. 173 


rst and fourth terms are called the extremes; and the second 
nd third terms, the means. 

Let a, }, c, and d, denote four quantities in arithmetical pro- 
ortion; and d the difference between either antecedent and its 
onsequent. 


Then, a—b=d, and a=b+d; 
also, e—d=d, and d=c—d. 


} 


By adding the last two equations, we have 
a+td=b-+e: that is, 


If four quantities are in arithmetical proportion, the sum of the 
‘v0 extremes is equal to the sum of the two means. 


| Arithmetical Progression. 


, 161. When the difference between the first antecedent and con- 
squent is the same as between any two consecutive terms of the 
roportion, the proportion is called an arithmetical progression. 
lence, an arithmetical progression, or a progression by differences, 
a succession of terms, each of which is greater or less than 
ie one that precedes it by a constant quantity, which is called 


ie common difference of the progression. Thus, 
Page 7 ID). 13; 16,19, 2%_:25; 5». 
and 60, 56, 52, 48, 44, 40, 36, 32, 28,... 


ve arithmetical progressions. The first is called an increasing 

‘ogression, of which the common difference is 3; and the sec- 

id, a decreasing progression, of which the common difference is 4. 

An arithmetical progression, is also called, an arithmetical series ; 

id generally, 

“A series is a succession of terms derived from each other accord- 
to some fixed and known law. 

I et a, b, c, d, e, f, . . . designate the terms of a progression 

differences ; it has been agreed to write them thus : 

es 

he a.b.c.d.e.f.g.h.t.k... 

} This series is read, a is to b, as b is to c, as ¢ is to d, as d 

‘toe, &c. This is 2 series of continued equi-differences, in which 








174 ELEMENTS OF ALGEBRA. [CHAP. VIL. 


each term is at the same time a consequent and antecedent, with 
the exception of the first term, which is only an antecedent, and 
_ the last, which is only a consequent. . 


162. Let d represent the common difference of the progression 
a.b.¢.¢.'f. 7a ere | 


which we will consider increasing. 
From the definition of a progression, it follows that, 


b=atd c=b+d=a42d, = 0 eee 


and, in general, any term of the series, is equal to the first term 
plus as many times the common difference as there are preceding 
terms. 

Thus, let 7 be any term, and m the number which marks the 
place of it. Then, the number of preceding terms will be deno- 
ted by » — 1, and the expression for this general term, will be 


l/=a+(n—1)d. 
That is, any term is equal to the first term, plus the product of 


the common difference by the number of preceding terms. 


If we make n=1, we have /=a; that is, the series will 
have but one term. 5 
If we make 


n= 2, we have 1=a+d; 


that is, the series will have two terms, and the second term is 
equal to the first plus the common difference. 


* 


EXAMPLES. 


/ 


1.1f a=3 and d=2, what is the 3d term? Ans. 7. 

2. If @a=5 and d=4, what is the 6th term? Ans. 25. 

3. If a=7 and d=5, what is the 9th term? Ans. ri | 

The formula, | | 
l=a+(n—1)d, 


serves to find any term whatever, without determining all those 
which precede it. 


y 
' 
co VII.] ARITHMETICAL PROGRESSION. 175 


Thus, to find the 50th term of the progression, 


fl 
ee 10.1316 19). 250. 


_we have, I=1+49 x 3= 148. 
And for the 60th term of the progression, 
Pe 13817 2 21...25, 4. 0, 
, we have, ~=1+59 x 4 = 237. 
163. If the progression were a decreasing one, we should have 


l= a— (n—1)d. 





i That is, any term in a decreasing arithmetical progression, is equal 
| the first term minus the product of the common difference by the 
umber of preceding terms. 

i 
| EXAMPLES. 


‘1. The first term of a decreasing progression is 60, and the 
ymmon difference 3: what is the 20th term? 





l=a—(n—1)d gives 1 = 60—(20 —1)3=60 — 57 =3. 


‘9. The first term is 90, the common difference 4: what is the 
5th term 2 Ans. 34. 





3. The first term is 100, and the common difference 2: what 
the 40th term ? Ans. 22. 


164. A progression by differences being given, it is proposed 
prove that, the sum of any two terms, taken at equal distances 
rom the two extremes, is equal to the sum of the two extremes. 

Let a.b.c.e.f....i.k.1, be the proposed progression, 


1d m the number of terms. 

We will first observe that, if « denote a term which has p 
rms before it, reckoning from the first term, and y a term which 
us p terms before it, reckoning from the last term, we have, 
om what has been said, 


and | y=l—pxd; 
whence, by addition, wx+y—=a+ ry 


176 ELEMENTS OF ALGEBRA. [CHAP. VIL 


Now, to find the sum of all the terms, write the progression 
below itself, but in an inverse order, viz., . 


ab. ¢. ef, oe 





Calling S the sum of the terms of the first progression, 2§ 


will be the sum of the terms in both progressions, and we shall 
have 


2S=(a+N)+ (bth +(c+i)...+ (ite) + (E+) +(1 +a). 


And, since all the ‘parts a+], b+ k, e-+%@.... are equal 
to each other, and their number equal to 2, by which we desig. 
nate the number of terms in each series, we have 





28 =(a+l1)n, or s= (45 


That is, the sum of the terms of an arithmetical progression, is 
equal to half the sum of the two extremes multiplied by the number 


of terms. 
EXAMPLES. 


1. The extremes are 2 and 16, and the number of terms 8: 
what is the sum of the series ? 








a+] , (OS eee tf 
| 5 ) xn, gives S= 5 x 8 = 72. 


2. The extremes are 3 and 27, and the number of terms 12: 
what is the sum of the series? Ans. 180. — 


3. The extremes are 4 and 20, and the number of terms 10: 
what is the sum of the series? Ans. 120. 


4. The extremes are 8 and 80, and the number of terms 10: 
what is the sum of the series? Ans. 440. 


165. The formulas 





Lina if, eee and s= (4) xs, 


contain five quantities, a, d, n, 1, and S, and consequently give 


rise to the following general problem, viz.: Any three of these 
five quantities being given, to determine the other two | 


JHAP. VII.] ARITHMETICAL PROGRESSION. es Wf 












































Given. a Values of the unknown Winntien ic] 
lS t=a-+ (n—1)d; S=4n[2a + (n — 1) d]. 
l—a (7+ a)(l1—a+d) 
aah = TOS WS a ee ee ee 
x ree Ss 2d 
d— d—2a+ (d—2a)?+8dS 
a,d,S| nj, tl |n= Seta S bi tant l=a+(n—1)d. 
S,d |S=4n(a+l); a dle Rs, 
n—]l 
an. iT a, t ee BAP 8"), plete a 
n(n — 1)’ n 
a,l,Si|n,d |n= a : salle ah ical oy 
a+l 28 — ee op + a) 





a, S |ja=l—(n—1)d; peer In [21 — - (n — 1) d]. 
28 —n(n—1)d_ uit 28 -+n(n—1)d 




















 Bid,n,S Gl hia = 





























2n 2n 
+: ] 2__ 8dS 
914,1,8| no [na tee Vv Cl+ar— sds, gaat (mt) dl 
2d | 
ON A? Sipe ole Ye aan 
10 |7,1, S| a, d ee uk 7299 — Cc ta y 
n n(n — 1) 


The solution of these cases presents no difficulty. Cases 3 and 
|) give rise to equations of the second degree; but one of the 
‘oots will always satisfy the enunciation of the question in its 
withmetical . sense. 


| air we resume the formula 

l=a+(n—1) d, 

| we have, a=1—(n—1)d; that is, 

| The first term of an increasing arithmetical progression, is equal 


}9 any following term, minus the product of the common difference 
ly the number of preceding terms. 


| From the same formula; we also find 
Har z 
Th l aa 


; 
n—l 


d — that is, 





12 


178 ELEMENTS OF ALGEBRA. [CHAP. VII 


In any arithmetical progression, the common difference is equal i 
the difference between the first and last terms considered, divided by 
the number of terms less one. 

1. Two terms of a progression are 16 and 4, and the numbe; 
of terms considered is 5: what is the common difference ? 

The formula 


(i= @ [6 2 
1 a ives” d= Sek 8) 
ep 4 
2. Two terms of a progression are 22 and 4, and the numbe: 
of terms considered is 10: what is the common difference ? 


Ans. 2: 


166. The last principle affords a solution to the followin; 
question :— , 

To find a number m of arithmetical means between two givei 
numbers a and b. 

To resolve this question, it is first necessary to find the com 
mon difference. Now we may regard a@ as the first term of ‘a! 
arithmetical progression, 6 as a subsequent term, and the require: 
means as intermediate terms. The number of terms of this pro 
gression which are considered, will be expressed by m + 2. 

Now, by substituting in the above formula, 6 for J, and m-+° 
for n, it becomes 


qo ee or Fibs! - 4 

m+2—1]1 

that is, the common difference of the required progression is ob 

tained by dividing the difference between the given numbers | 
and 6, by one more than the required number of means. 

Having obtained the common difference, form the second terr 
of the progression, or the first arithmetical mean, by adding d, 0 
b—a 
ie ae 
menting the first by d, &c. 


to the first term a. The second mean is obtained by aug 


1. Find 3 arithmetical means between 2 and 18. The formul 


nence the progression is 
2.6.10.14.18. 


)tAP. VII.) ARITHMETICAL PROGRESSION. VTS 


2. Find 12 arithmetical means between 12 and 77. The for- 
ila | 
b—a ; 77 — 12 
dies ETA ih Sn Cie 
eG he gives d 3 5; 
nce the progression is 
EOE Y Ep Ey ie aa [Os ee 


Wee. Remarc. —If the same number of arithmetical means are 
serted between the terms of a progression, taken two and two, 
ase terms, and the arithmetical means united, will form one and 
» same progression. 


Mei set a. Ob oc.e ff... 





. be the proposed progression, and 
the number of means to be inserted between a and b, 6 and ce, 





/From what has just been said, the common difference of each 
‘ttial progression will be expressed by 

' b—a c—b e—c 

m+1*> m+1’ m+1 

uch are equal to each other, since a, b, c, ... are in pro- 
'3ssion: therefore, the common difference is the same in each 
the partial progressions ; and since the last term of the first, 
'ms the first term of the second, d&c., we may conclude that 
| of these partial progressions form a single progression. 


EXAMPLES. 
| 


1. Find the sum of the first fifty terms of the progression 
: MeegeetO..23 .. . 

For the 50th term, we have 

tee. ; Pee 2 4649 x 7 = 345. 

Ve 50 

Hence, S= (2 -+- 345) x % = 347 x 25 = 8675. 

(2. Find the 100th term of the series 2.9.16.23... 


Ans. 695. 
\3. Find the sum of 100 terms of the series 1.3.5.7.9... 
Ans. 10000. 
t. The greatest term considered is 70, the common difference 
and the number of terms 21: what is the least term and the 
a of the series ? 


Ans. Least term 10; sum of series 840. 





180 ELEMENTS OF ALGEBRA. (CHAP. VII, 


5. The first term of a decreasing arithmetical progression is 
10, the common difference one third, and the number of terms 
21: required the sum of the series. * Ans. 140. 


6. In a progression by differences, having given the common 
difference 6, the last term 185, and the sum of the terms 2945; 
tind the first term, .and the number of terms. 

Ans. First term =5; number of terms 31. 


7. Find 9 arithmetical means between each antecedent and cop- 
sequent of the progression 2.5.8.11.14... 
Ans. d= 0.3; 
8. Find the number of men contained in a triangular battalion, 
the first rank containing 1 man, the second 2, the third 3, and 
s0 on to the n‘*, which contains 7. In other words, find the ex- 





pression for the sum of the natural numbers 1, 2, 3, .. . from 

1 to n, inclusively. Vere n(n 7 
9. Find the sum of the n first terms of the progression of wr 

even vnumpers., 15°3)°6, 7,9. 9% Ans. S= 74 


10. One hundred stones being placed on the ground, in a 
straight line, at the distance of 2 yards from each other, how 
far will a person travel, who shall bring them one by one to 4 
basket, placed at two yards from the first stone ? 

Ans. 11 miles, 840 yards. 


Geometrical Proportion. 


168. Ratio is the quotient arising from dividing one quantity 
by another quantity of the same kind. Thus, if A and B repre 
sent quantities of the same kind, the ratio of A to B is expresset 
by 


— 
e 


A 


169. If there be four magnitudes, A, B, C, and D, having sucl 
values that 


aC 
then A is said to have the same ratio to B, that C has to D 
or, the ratio of A to B is equal to the ratio of C to D. Whe! 


i} 
i 
i) 


HAP. VII.] GEOMETRICAL PROPORTION. 181] 





mur quantities have this relation to each other, they are said to 
\2 in proportion. Hence, proportion is an equality of ratios. 

To express that the ratio of A to B is equal to the ratio of C 
|: D, we write the quantities thus, 


pemtegeriee te Coed, 


| The Aianition Pick are serpered together are called the terms 
“the proportion. ‘The first and last terms are called the two ex- 
| emes, and the second and third terms, the two means. 


170. Of four proportional quantities, the first and third are called 
|e antecedents, and the second and fourth the consequents ; and 


e last is said to be a fourth proportional to the other three 
ken in order. 














W 171. Three quantities are in proportion when the first has the 
‘me ratio to the second that the second has to the third; and 
en the middle term is said to be a mean proportional a 
.e other two. 


| 172. Quantities are said to be in proportion by inversion, or in- 
rsely, when the consequents are made the antecedents and the 
jitecedents the consequents. 

i 


173. Quantities are said to be in proportion by alternation, or 
Ae ‘ ° 
| ternately, when antecedent is compared with antecedent and con- 


quent with consequent. 





| 174. Quantities are said to be in proportion by composition, when 
;e sum of the antecedent and consequent is compared either with 


; 


\tecedent or consequent. 


| 175. Quantities are said to be in proportion by division, when 
ye difference of the antecedent and consequent is compared either 
: th antecedent or consequent. 


176. Equi-multiples of two or more quanatier are the products 
uch arise from multiplying the quantities by the same number. 
jaus, mx A and mx B, are equi-multiples of A and B, the 
| mmon multiplier being m. 


177. Two quantities, A and B, are said to be reciprocally pro- 
rttonal, or inversely proportional, when one increases in the same 
tio as the other diminishes. When this relation exists, either 
them is equal to a constant quantity divided by the other 


182 ELEMENTS OF ALGEBRA. [CHAP. VI 


178. If we have the proportion 


Ay Been ay: * 
B D 
we have ot ay (Art. 169); 


and by clearing the equation of fractions, we have 
BC = AD; that is, 
Of four proportional quantities, the product of the two extrem 
is equal to the product of the two means. 
179. If four quantities, A, B, C, and D, are so related to ear 
other that 
Arid secs 
BAe 
Ul h — = 
we shall also have, n a 
and hence} A: Buty Cos DY thatts 
If the product of two quantities is equal to the product of ti 
other quantities, two of them may be made the extremes, and t 
other two the means of a proportion. 
180. If we have three proportional quantities, 
A’: Bis 20 B. ae 


we hav Pte &. 
s Ae ae 
hence, B? = AC; that is, 


The square of the middle term is equal to the product of the t 


extremes. 


181. If we have 


A: B:: C: D, and consequently, 4-2, 


multiplying both members of the equation by “A we obtain | 
oO D | 
Be Be 
and hence, A: OF: BB: Ds sie 


If four quantities are proportional, they will be in proportion 
alternation. 










‘HAP. VII.) GEOMETRICAL PROPORTION. 183 


182. If we have ! 

‘© Beer ee: 2). and ApshBaw: fh tubs 

‘ve shall also have 

B D 
re 


—_— — 


and cide ipa 
A Agicuk + 


hence, 


== aida) > ch © Ps that is, 


Q|s 


_ Of there are two sets of proportions having an antecedent and con 
requent in the one equal to an antecedent and consequent of the 
ther, the remaining terms will be proportional. 


ts 183. If we have 


Boa(P 

Bat 25> C.: D, and consequently, ry es 
we have, by dividing 1 by each member of the equation, 
iP aec 
B= D 
| Four proportional quantities will bein proportion, when taken wm- 
versely (Art. 172). 
184. The proportion 
| meets  ; D, gives AX P= BX. C. 

To each member of the last equation add B x D. We shall 
‘hen have 


and consequently, B: A:: D: C; that is, 


(A+ B)x D=(C+D)xB; 
and by separating the factors, we obtain 
Ae pe b:: C+D": D. 
If, instead of adding, we subtract B x D from both members, 
we have 


(A —B)x D=(C—D) xB; 
lwhich gives A—B:B:: C—D: D; that is, 


If four quantities are proportional, they will be in proportion by 
fsomposition or division. 


185. If we have 


Q|o 


B 
A 


? 


184 / ELEMENTS OF ALGEBRA. [CHAP. Vit. 


and multiply the numerator and denominator of the first member 
by any number m, we obtain a 


ple kame bY 
ey € 
Equal multiples of two quantities have the same ratio as the quan- 


and mA: mB: : C.: D; that is, 


tities themselves. 
186. The proportions 
A? B:? CD, “and Ales = eee 
give AxD=8B.x C, and ,.Ax® Ff =e 
adding and subtracting these equations, we obtain 
A(D+F)=B(C+#E), or A: B::CHE: D+F; that ia, 


If C and D, the antecedent and consequent, be augmented or 
diminished by quantities E and F, which have the same ratio as C 





to D, the resulting quantities will also have the same ratio. 
187. If we have several proportions, 


A:B::C:D, which gives AxD=BxC, 


MEET 8 Bek By faa oF ce fs * mex FoBxX E, 
Aye ist ce Eee .e 6 Mx. HSB G : 
&c., &c., 


we shall have, by addition, 
A(D+ F+ H)=B(C+ E+ G); 
and by separating the factors, 
A: B:C+E+G: D+F+4+4; that is, 
In any number of proportions having the same ratio, any antece- 


dent will be to its consequent, as the sum of the antecedents to the 
sum of the consequents. 


188. If we have four proportional quantities 


As) Bu Cogs have asi 
and raising both members to any power, as the ath, we have : 
Bien Le | 
> Ge 


and consequently, Ar ; B®: : C* : Des thei 


HAP. VII.) GEOMETRICAL PROGRESSION. 185 


“Tf four quantities are proportional, any like powers or roots will 
|? proportional. 


"189. Let there be two sets of proportions, 


A:B::C: D, which gives 
Pea Gi ss gs 


| Multiply them together, member by member, we have 


Mo DIS: 0. 
: aieres which gives AE: BF:: CG: DH; that is, 
‘In two sets of proportional quantities, the products of the corres- 
nding terms will be proportional. 





Of Geometrical Progression. 





190. In the proportions which have been considered, it has 
jily been required that the ratio of the first term to the second 
jould be the same aswthat of the third to the fourth. If we im- 
ise the farther sonal that the ratio of the second to the 
|ird shall also be the same as that of the first to the second, or 
| the third to the fourth, we shall have a series of numbers, 
ch of which, divided by the preceding one, will give the same 
itio. Hence, if any term be multiplied by this quotient, the 
oduct will be the succeeding term. A series of numbers so 
rmed is called a geometrical progression. Hence, 

| A geometrical progression, or progression by quotients, is a series 
, terms, each of which is equal to the product of that which 
jecedes it by a constant number, which number is called the 
tio of the progression. Thus, in the two series, 


3, 6, 12, 24, 48, 96, 





64, 16, 4, 1, 


ch term of the first contains that which precedes it twice, or 
equal to double that which precedes it; and each term of the 
cond contains the term which precedes it one-fourth times, or 
a fourth of that which precedes it. These are geometrical pro- 


186 ELEMENTS OF ALGEBRA. [CHAP. Vil. | 


gressions. In the first, the ratio is 2; in the second, it is 4. 


The first is called an increasing progression, the second a de- 


\ 


creasing progression. 


Let a, b, c, d, e, f,; . . . denote numbers in a progression by 
quotients: they are written thus: 


@: 0b: ¢ ta: 6 See 

and it is enunciated in the same manner as a progression by dif: 
ferences. It is necessary, however, to make the distinction, that 
one is a series of egual differences, and the other a series of 
equal quotients or ratios. It should be remarked, that each term 
of the progression is at the same time an antecedent and a con- 
sequent, except the first, which is only an antecedent, and the 
last, which is only a consequent. 


191. Let r denote the ratio of the progression 
aveod': occ eyiRaae 


r being >1 when the progression is increasing, and r<1 when 
it is decreasing. We deduce from the, definition, the following 


equations : : 
b= ar, (c= br = ar, nt os ae a 


and, in general, any term 7, that is, one which has n — 1 terms 
before it, is expressed by ar™—'. 

Let 7 be this term; we have the formula 

beshonnty t 

by means of which we can obtain any term without being obliged 

to find all the terms which precede it. That is, | 


Any term of a geometrical progression is equal to the first tern 
multiplied by the ratio raised to a power whose exponent denote 
the number of preceding terms. 


EXAMPLES. 


1. Find the 5th term of the progression 
: 2isyv4d se8 >) 16; sheeg T 
in which the first term is 2, and the common ratio 2. 


5th term =2 x 24=2 x 16 = 32. 


i 









-QUHIAP. VII.) GEOMETRICAL PROGRESSION. 187 


2. Find the 8th term of the progression 
Bee sO 2) OS he ag 

Sib term eee. x. 3? = 2.x 2187 = 4374, 
3. Find the 12th term of the progression 


cote ae ea ea 
4 
11 3 
Paha ni cie=-64 Ge) ar ST Re a 
4 4il 48 65536 


192. We will now explain the method of determining the sum 
of n terms of the progression 


PUG Gite eee 208s ot gs le Be 
‘of which the ratio is r. _ 


_ If-we denote the sum of the series by S, and the nth term 
by J, we shall have 
3 S=atar+ar.... tar? + ar), 
___ If we multiply both members by 7, we have 

Sr = ate ar? tar... +ar*!+ ar; 
and by subtracting the first equation, 


ar™ —a 
> 





Sr — S = ar™* — a, whence, S= 
9 ’ r 1 


and by substituting for ar”, its value lr, we have 


if — 2 


ie 





pee Lis 


. That is, to obtain the sum of any number of terms of a pro- 
| gression by quotients, multiply the last term by the ratio, subtract 
the first term from this product, and divide the remainder by the 
ratio diminished by unity. 


EXAMPLES. 


1. Find the sum of eight terms of the progression 
2:6: 18.; 6439162 ...: 2.x 3? = 4374. 
lr—a 13122 — 2 


= 5000. 


r—l 


188 ELEMENTS OF ALGEBRA. (CHAP. VII. 


2. Find the sum of five terms of the progression 
2 > 4 2° Bal ibe: eee 





lr—a 64—2 
= = 62. 
r— 1 1 
3. Find the sum of ten terms of the progression 
2:6: 18: 54: 162 1. . go? eas 


Ans. 59048. 
4. What debt may be discharged in a year, or twelve months, 
by paying $1 the first month, $2 the second month, $4 the third 
month, and so on, each succeeding payment being double the last; 
and what will be the last payment? ; 
Ans. Debt, $4095; last payment, $2048. 


5. A gentleman married his daughter on New-Year’s day, and 
gave her husband ls. toward her portion, and was to double it 
on the first day of every month during the year: what was her 
portion ? ; Ans. £204 15s. 


6. A man bought 10 bushels of wheat on the condition that he 
should pay 1 ‘cent for the first bushel, 3 for the second, 9 for 
the third, and so on to the last: what did he pay for the last 
bushel and for the ten bushels? ® 

Ans. Last bushel, $196,83; total cost, $295,24. 

193. When the progression is decreasing, we have 7 < 1 and 
1< a; the above formula for the sum is then written under the 
form : 





in order that the two terms of the fraction may be positive. 
By substituting ar™~! for 7 in the expression for S, it becomes 





EXAMPLES. 
1. Find the sum of the first five terms of the progression | 
$2 :\16 28 : 4 2 2 





i 
eos | ogee a aie 
ra ey ae | | ee 
2 2 


JHAP. VII] GEOMETRICAL PROGRESSION. 189 


_ 2. Find the sum of the first twelve terms ‘of the progression 





a SR ee : 64(<) , o : 
" ; : : $ neu = 2 sag 4 ; De 65536" 
1 1 1 
PMR Tope © 
gan 65536 <4 ue 6 — §5536 Ay 65535 
(parties 3 rm Sil) 196608" 


_ We perceive that the: principal difficulty consists in obtaimmg 
he numerical value of the last term, a tedious operation, even 
vhen the number of terms is not very great. 


194. Remarx.—lIf, in the formula 


we suppose r = 1, it becomes 


0 
Sait 


This result is a symbol of indetermination. It often arises 
‘rom the existence ofa common factor (Art. 113), which becomes 
iothing by making a particular hypothesis on the quantities which 
mter the equation. If this common factor can be divided out, the 
Xpression will assume a determinate form. This, in fact, is the 
‘ase in the present question; for, the expression r* — 1 is divisi- 
le by r—1 (Art. 61), and gives the quotient 


gel to pn—2 ten 3 tf ter tl; 
tence, the value of S takes the form 
S = ar™! + art-2 + ar™3 + 2... fart+a. 
Now, making r=1, we have 
S=a+t+at+a+...+t+a=na. 


We can obtain the same result by going back to the proposed 
Togression 





b @eedecdy is «02 G, 





190 . ELEMENTS OF ALGEBRA. (CHAP. VII 


The result x given by the formula, may be regarded as in- 


dicating that the series is characterized by some particular prop- 
erty. In fact, the progression, being entirely composed of equal 
terms, is no more a progression by quotients than it is a pro- 
gression by differences. Therefore, in seeking for the sum of a 
certain number of the terms, there is no reason for using the 
formula 











which gives the sum in the progression by differences. 


195. The consideration of the five quantities, 4, r,n, 1, and Sj 
which enter into the formulas 


/=ar*-! and S= i a 
r—1 
give rise to several curious problems. . : 

Of these cases, we shall consider here, only the most im- 
portant. We will first find the values of S and r in terms of 
Gawis ana fe 


The first formula gives 


1 | are A 
r—-l1 — —,. whence r= —. 
a a 


Substituting this value in the second formula, the value of S will | 
be obtained. 


The expression 
n—1 7 
r= ae 


a 


> 


furnishes the means for resolving the following question, viz.: 

To find m mean proportionals between two given numbers a and 
b; that is, to find a number m of means, which will form with a } 
and b, considered as extremes, a progression by quotients. | 

To find this series, it is only necessary to know the ratio. 
Now, the required number 2f means being m, the total number 











- 


MAP. VII.] GEOMETRICAL PROGRESSION. 19] 


terms considered, will be equal to m-+ 2. Moreover, we have 
=, therefore the value of r becomes 


m+1 i 
| Sm 9 
a 


jat is, we must divide one of the given numbers (b) by the other 
:), then extract that root of the quotient whose index ts one more 
an the required number of means. 

‘Hence, the progression is 


1 imc 2 | — .@4 ooo te Ch eee PRCT ) 

a a as 
Thus, to insert six mean proportionals between the numbers 
‘and 384, we make m= 6, whence 


" / 384 
| r= oo, 128 = 2; 


: 
/ 


hence we deduce the progression 
peo ten oe 2 46": 96 F192") 384: 


'Remarx.—When the same number of mean proportionals are 
iserted between all the terms of a progression by quotients, taken 
vo and two, all the progressions thus formed will constitute a 
ngle progression. 


Of Progressions having an infinite Number of Terms. 


196. Let there be the decreasing progression 
COPD a Ct I a a 
mtaining an indefinite number of terms. The formula 


: ; a— ar* 


5 





Now, since the progression is decreasing, r is a proper frac- 
on, and r” is also a fraction, which diminishes as n increases. 
‘herefore, the greater the number of terms we take, the more 


d a ark 
rill x; x r® diminish, and consequently, the more will the 
: eal is 


192° ELEMENTS OF ALGEBRA. (CHAP. VII 


partial sum of these terms approximate to an equality with the 





first part of S; that is, to Finally, when n is taken 


—?T 
greater than any given number, or 


<r 





m=, then 
—r 


will be less than any given number, or will become equal to 0; 


and the expression will represent the-true value of the 





sum of all the terms of the series. 


Whence, we may conclude, that the expression for the sum of 
the terms of a decreasing progression, in which the number of terms 
ts infinite, 1s 

a 
S's Hee 

This is, properly speaking, the limit to which the partial sums 
approach, by taking a greater number of terms of the progression. 
The number of terms may be taken so great as to make the dif 


Chg a 
ference between the sum, and aie as small as we please, and 


the difference will only become nothing when the number of term} 
taken is infinite. | 


EXAMPLES. 


1. Find the sum of 
1 1 1 1 Nag rg 
3 : 9 4 37 : 81 to infinity. 
We have, for the sum of the terms, 
a 1 3 


l—r 1 2° 
3 





2. Again, take the progression 
si 1 1 1 1 
my : ay : 3 : 16 ° 32 ee CP > 


WwW = SC 
e have S ae : 
foe 

3 


ay 





What is the error, in each example, for n= 4, a=5, n= 6! 


fof lewrhseas 
i § 


‘MAP. VIIT.] PERMUTATIONS AND COMBINATIONS. 193 


‘' 





\ 


CHAPTER VIII. 


} 
i 
ORMATION OF POWERS, AND EXTRACTION OF ROOTS OF ANY DEGREE. 
j —CALCULUS OF RADICALS.—INDETERMINATE CO-EFFICIENTS. 

| 197. The resolution of equations of the second degree supposes 
'e process for extracting the square root to be known. In like 
anner, the resolution of equations of the third, fourth, &c. de- 
Yee, requires that we should know how to extract the third, 
arth, &c. root of any numerical or algebraic quantity. 

|The power of a number can be obtained by the rules of mul- 
lication. and this power is subjected to a certain law of forma- 
im, which it is necessary to know, in order to deduce the root 
om the Power: 

“Now, the law of formation of the square of a numerical or 
zebraic quantity, is deduced from the expression for the square 
a binomial (Art. 116); so likewise, the law of a power of any 
gree, is deduced from the same power of a binomial. We shall 
srefore first determine the development of any power of a binomial. 








198. By multiplying the binomial «+a into itself several times, 
2 following results are obtained: 

(2 --\ 4), =a 2, 

(2 + a)? = a? + 2ax + a?, 

(a + a)? = a? + 3ax? + 30x + a’, 

(# + a)/* = xt + 4ax° + 6a?x? + 4a3x + a’, 

(a + a) = x + 5ax* + 10a7x? + 10a3x? + Sate + a’. 
By examining the developments, we readily discover the law 


cording to which the exponents of x decrease and those of @ 
Tease, in the successive terms; it is not, however, so easy to 
. 13 


194 ELEMENTS OF ALGEBRA. [CHAP. VIIT 


discover a law for the co-efficients. Newton discovered one, by 
means of which a binomial may be raised to any power, without! 
first obtaining all of the inferior powers. He did not, however 
explain the course of reasoning which led him to the discovery 
but the law has since been demonstrated in a rigorous manner 
Of all the known demonstrations of it, the most elementary is 
that which is founded upon the theory of combinations. However 
as the demonstration is rather complicated, we will, in order t 
simplify it, begin by resolving some problems relative to permuta 
tions and combinations, on which the demonstration of the formnul: 
for the binomial theorem depends. 


Theory of Permutations and Combinations. 


199. Let it be proposed to determine the whole number of way 
in which several letters, a, 6, c, d, &c., can be written one afte 
the other. The result corresponding to each change in the posi 
tion of any one of these letters, is called a permutation. 

Thus, the two letters a and 6 furnish the two permutations, a 
and ba. 


; abe 

ach 

In like manner, the three letters, a, 5, c, furni a cab 
six permutations. bac 
ay 

cha 


Permutations, are the results obtain 1 by y writing a certain nun 
ber of letters one after the other, in every possible order, in such 
manner that all the letters shall enter into each result, and au 
letter enter but once. oy 


Rw a 


Prositem 1. To determine the number of permutations of whic 
n letters are susceptible. 


In the first place, two letters, a and b, evidently ah 
give two permutations. ; ba 
Therefore, the number of permutations of two letters is «4 
pressed by 1 x 2. | 
Take the three letters, a, 5, and c. Reserve c 
either of the letters, as c, and permute the other ab 
two, giving be 


di 















‘gap. VIII.] PERMUTATIONS AND COMBINATIONS. 195 


» Now, the third letter c may be placed before ab, cab 
xtween a and 0, and at the right of ab; and the 
me for ba: that is, in ONE of the first permuta- 
uns, the reserved letter ¢ may have three different abe 
aces, giving three permutations. Now, as the same : Sa 
jay be shown for each one of the first permutations, | - 
; follows that the whole number of permutations of 
ree letters will be expressed by, 1 x 2 x 3. bac 
| If now, a fourth letter d be introduced, it can have four places 
4 each one of the six permutations of three letters: hence, all 
| € permutations of four letters will be expressed by, 1 x2 x3 x 4. 
' In general, let there be n letters, a, 6, c, &c., and suppose the 
tal number of permutations of n— 1 letters to be known; and 
t Q denote that number. Now, in each one of the Q permuta- 
‘ms, the reserved letter may have n places, giving n permutations : 
‘mee, when it is so combined with all of them, the entire num- 
‘r of permutations will be expressed by Q x n. 

Let n=2. Qvwill then denote the number of permutations that 
in be made with a single letter; hence, Q=1, and in this 
| rticular case we have, Q x n=1 x 2. 

‘Let n=3. Q will then express the number of permutations 
'3—1 or 2 letters, and is equal to 1 x 2. Therefore, Q x n 
t equal to 1 x 2 x 3. 

Let n= 4. Q in this case denotes the number of permutations 
' 3 letters, and is equal to 1 x 2 x 3. Hence, Q x n becomes 
|X 2x 3 xX 4; and similarly, when there are more letters. 





200. Suppose we have a number m, of letters a, 6, c, d, &c. 
| they are written one after the other, in classes of 2 and 2, or 
‘and 3, or 4 and 4... . in every possible order in each class, 
such a manner, however, that the number of letters in each 
sult shall be less than the number of given letters, we may de- 
md the whole number of results thus obtained. These results 
‘ie called arrangements. 

| Thus, ab, ac, ad, . . . ba, be, bd, . . . ca, cb, ed, . . . are ar- 
ngements of m letters taken 2 and 2; or in sets of 2 each. 
In like manner, adc, abd, Prodc, ‘OGG, > sh gtr, OChwineerr are 
rangements taken in sets of 3. 

Arrangements, are the results obtained by writing a number m of 
ters one after the other in every possible order, in sets of 2 and 


196 ELEMENTS OF ALGEBRA. [CHAP. Vuit. 


2, 3 and 3, 4 and 4...n andn; m being >n; that is, the 
number of letters m each set being less than the whole numbe 
of letters considered. If, however, we suppose n =m, the ar 
rangements taken n and n, will become simple permutations. 


ProsLEem 2. Having given a number m of letters a,b,c, d . . 
to determine the total number of arrangements that may be formed of 
them by taking them n at a time; m being supposed greater than n 

Let it be proposed, in the first place, to arrange the three let 
ters, a, 6, and c, in sets of two each. 

First, arrange the letters in sets of one each, and for 


a 
each set so formed, there will be two letters reserved: ; 
the reserved letters for either arrangement, being intl 

which do not enter. ¢ 


When we arrange with reference to a, the reserved letters wil 
be 6 and c; if with reference to 6, the reserved letters will he 
a and c, &c. 


Now, to any one of the letters, as a, annex, in suc- y 
cession, the reserved letters 6 and c: to the second | 1 
arrangement b, annex the reserved letters @ and c; and r 
to the third arrangement, c, annex the reserved letters J 
a and d: this gives ; Lol 


And since each of the first arrangements is repeated as many 
times as there are reserved letters, it follows, that the arrange 
ments of three letters taken two in a set, will be equal to the of 
rangements of the same number of letters taken one in a set, multi 
plied by the number of reserved letters. 

Let it be required to form the arrangement of four leas 
a, 6, c, and d, taken 3 in a set. , 


First, arrange the four letters in sets of two: there a 
will then be two reserved letters. Take one of the \ 
sets and write after it, in succession, each of the re- ‘La 
served letters: we shall thus form as many sets of ie 
three letters each as there are reserved letters; and these 7 


sets differ from each other by at least the last letter. 
Take another of the first arrangements, and annex in 


In sets of two 
) Bom, Sth Re 


succession the reserved letters; we shall again form he 
as many different arrangements as there are reserved ; 
letters. Do the same for all of the first arrangements, | 
and it is plain, that the whole number of arrangements 




















NHAP. VIII.) PERMUTATIONS AND COMBINATIONS. 197 


“hich will be formed, of four letters, taken 3 and 3, will be equal 
» the arrangements of the same letters, taken two in a set, mul- 
plied by the number of reserved letters. 

In order to resolve this question in a general manner, suppose 
ie total number of arrangements of m letters, taken n —1 in a 
+t, to be known, and denote this number by P. 

Take any one of these arrangements, and annex to it, in suc- 
sssion, each of the reserved letters, and of which the number 
_m—(n-—-1), or m—n+1: it is evident, that we shall thus 
‘rm a number m—n-+1 of new arrangements of n letters, each 
ffering from the other by the last letter. Now, take another of the 
‘st arrangements of n — 1 letters, and annex to it, in succession, 
vch of the m—n-+ 1 letters which do not make a part of it; we 
a obtain a number m—n-+ 1 of arrangements of 7 letters, 
| flering from each other, and from those obtained as above, at least 
one of the » —1 first letters. Now, as we may in the same 
anner, take all the P arrangements of the m letters, taken n — | 
a set, and annex to each in succession each of the m—n-+1 
her letters, it follows that the total number of arrangements of 
‘letters taken n in a set, is expressed by 


P(m—n + 1). 


| To apply this in determining the number of arrangements of 
| letters, taken 2 and 2, 3 and 3, 4 and 4, or 5 and 5 in a set, 
‘ake nm=2; whence, m—n+1—=m—1; P in this case, 
ill express the total number of arrangements, taken 2 —1 and 
'—1, or 1 and 1; and is consequently equal to m; therefore, 
le formula becomes m (m — 1). 

| Let n»=3; whence, m—n+1—=m—2; P will then ex- 
jess the number of arrangements taken 2 and 2, and is equal to 
\(m—1); therefore, the formula becomes | 


> 


m(m — 1)(m — 2). 


| Again, take n= 4: whence, m—n+1=m—3: Pvwill ex- 
. 


jess the number of arrangements taken 3 and 3, or is equal to 


m(m — 1) (m — 2); + Ag, 
orefore, the formula becomes 


m(m — 1) (m — 2) (m — 8). m.* 


198 ELEMENTS OF ALGEBRA. [CHAP. VIL 


Remark.—From the manner in which these results have bee 
deduced, we conclude that the general formula for m letters take 
n in a set, is . 

m (m — 1)(m — 2)(m — 3)... . (m—n+$ 1); 


that is, it is composed of the product of the n consecutive number 
comprised between m and m—n + l, inclusively. 

From this formula, that of the preceding Art. can easily } 
deduced, viz., the development of the value of Q X n. 

For, we see that the arrangements become permutations whe 
the number of letters entering into each arrangement is equal | 
the total number of letters considered. 

Therefore, to pass from the total number of arrangements of 
letters, taken n and n, to the number of permutations of 7 letter 
it is only necessary to make m= in the above developmer 
which gives 

n(n — 1) (n—2)(n—3) ~~... i 


By reversing the order of the factors, and observing that tl 
last is 1, the next to the last 2, the third from the last 3, & 
we have F 


for the development of Q xX 2. 
This is nothing more than the series of natural numbers cor 


prised between 1 and a, inclusively. 


201. When the letters are disposed, as in the arrangements, 
and 2, 3 and 3, 4 and 4, &c., it may be required that no two | 
the results, thus fotmed, shall be composed entirely of the san 
letters, in which case the products of the letters will be differen 
and we may then demand the whole number of results thus 0 
tained. In this case, the results are called combinations. 


Thus, ab, ac, be, . . . ad, bd, . . . are combinations of the kk 
ters a, b, and c, &c., taken 2 and 2. 
In like. manner, abc, abd, . . . acd, bed, . . . are combinatio 


of the letters taken 3 and 3. 

ee are arrangements in which any two will differ fre 
each other by at least one of the letters which enter them. 
Hence, there is an essential difference in the signification | 
the words, permutations, arrangements, and combinations. 








HAP. VIIL.] PERMUTATIONS AND COMBINATIONS. 199 


“Prosiem 3. To determine the total number of different combina- 
ms that can be formed of m letters, taken n in @ set. 


‘Let X denote the total number of arrangements that can be 
rmed of m letters, taken x and n: Y the number of permuta- 
ms of n letters, and Z the total number of different combinations 
‘ken n and n. 

It is evident, that all the possible arrangements of m letters, 
ken n in a set, can be obtained, by subjecting the n letters of 
ach of the Z combinations, to all the permutations of which 
‘ese letters are susceptible. Now, a single combination of x let- 
‘rs gives, by hypothesis, Y permutations ; therefore Z combina- 
ons will give Y x Z arrangements, taken .. and 7; and as X 
snotes the total number of arrangements, it follows that the three 
jantities, X, Y, and Z, give the relations 





X=YxZ; whence, Za 
But we have (Art. 200), 
X= P(m—n+ 1), 
and (Art. 199), Y=Qx2n; 
therefore, Z=— > vrata Bala x sidbesihs Jae 
<n Q n 


‘Since P- expresses the total number of arrangements, taken 
—1 and n—1, and Q the number of permutations of n—1 


| i 
‘itters, it follows that @ expresses the number of different com- 


nations of m letters taken n —1 and n — 1.’ 
To apply this to the case of the combinations of m letters 


ken 2 and 2, 3 and 3, 4 and 4, &c. 
Make n = 2, in which case, 5 expresses the number of com- 


mations of m letters taken 2—1 and 2—1, or taken 1 and 1, 
ad this number must be equal to m; the above formula there- 
re becomes 


PDS Bs wi te} 
Let n= 3; Q will express the number of combinations taken 


200 ELEMENTS OF ALGEBRA. * "CHAP. VIET. 








— 1 
2 and 2, and is equal to _ 5 ita: and the formula becomes 
m {mn Ve ae 
123 


In like manner, we find the number of combinations of 2 let- 
ters taken 4 and 4, to be 
m (m — 1) (m —2) (m— 3) | 
112.3148 f 





and, in general, the number of combinations of m letters taxen n 
and n, is expressed by 








m(m — 1) (m — 2) (m — 3) (m—n-+ 1) 
1.2:3:4,104. Na ree ae 
which is the development of the expression 
P(m—n+1) 
Qxn° 


We may here observe that, if we have a series of numbers, 
decreasing by unity, and of which the first is m and the last 
m—p, m and p being entire numbers, that the product of these 
numbers will be exactly divisible by the continued product of all 
the natural numbers from 1 to p-+ 1 inclusively; that is, 


m(m — 1) (m— 2) (m— 3)... (m—p) 
1. 2. es oe 


is a whole number. For, from what has been proved, this ex- 
pression represents the number of different combinations that can 
be formed of m letters taken in sets of p+1 and p+1. Now 
this number of combinations is, from its nature, an entire number; 
therefore the above expression is necessarily a whole number. 





Demonstration of the Binomial Theorem. 


202. In order to discover more easily ‘the law for the develop- 
ment of the mth power of the ‘binomial « + a, let us observe — 
the law of the product of several binomial factors, ‘at +a, r+, 
x-+c, x+d... of which the first term is the same in each, 
and the second terms different. 


4 








KAP. VIII.) | * BINOMIAL THEOREM. 201 
} x«+a 

| za2+ob 

fh product - «2+ a | x + ab 

+ b| 

’ e+e 

| Sc +a] a2?+ ab|a« + abe 

| +6| + ae 

+ ¢ + be 

x-+d 

i - - = at + oi ab | x? + arg + abcd 
{ + b | +ac, + abd 

| +ec| + ad + acd 

| DA of ho... bad 

h + bd 

+ cd 





fe sé 


| 
_ From these products, obtained by the common rule for alge- 
‘Taic multiplication, we discover the following laws :— 


L Ist. With respect to the exponents, we observe that, the ex- 
sonent of x, in the first term, is equal to the number of binomial 
actors employed. In each of the following terms to the right, 
‘dis exponent diminishes by unity to the last term, where it is 0. 


2d. With respect to the co-efficients of the different powers of 
', that of the first term is unity; the co-efficient of the second 
‘rm is equal to the sum of the second terms of the binomials ; 
ae co-efficient of the third term is equal to the sum of the prod- 
‘ets of the different second terms, taken two and two; the co- 
ficient of the fourth term is equal to the sum of their different 
roducts, taken three and three. Reasoning from analogy, we may 
onclude that the co-efficient of the term which has n terms be- 
wre it, is equal to the sum of the different products of the second 
ms of the m binomials, taken n and n. The last term of the 
roduct is equal to the continued product of the second terms of 
qe binomials. 

In order to prove that this law of formation is general, suppose 
iat it has been proved true for a number m of binomials ; let us 





202 ; ELEMENTS OF ALGEBRA. ”  [CHAP. VIII. 


see if it will continue to be true when the product is multiplied 
by a new factor. 
For this purpose, suppose 


a+ Avm—1 4. Bam—2+ Cam3 ... + Mar—t1 4+ Naman+t ...+0U, 
to be the product of m binomial factors, Na"~”" representing the 


term which has n terms before it, and Mx™-"+! the term which 
immediately precedes. 

Let «+ k be the new factor by which we multiply ; the prod- 
uct when arranged according to the powers of a, will be 


omtit Alamt B lam1+C |am24+...4N |am-ati+ .-, 

+ k + Ak | + Bk + Mk + Uk, 

From which we perceive that the law of the exponents is evi 
dently the same. 


With respect to the co-efficients, we observe, 

Ist. That the co-efficient of the first term is unity; and 

2d. That A-+&, or the co-efficient of a”, is the sum of the 
second terms of the m+ 1 binomials. 

3d. Since, by hypothesis, B is the sum of the different products 
of the second terms of the m binomials, taken two and two, and 
since A xX k expresses the sum of the products of each of the 
second terms of the m binomials by the new second term 4; there- 
fore, B+ Ak is the sum of the different preducts of the second 
terms of the m+ 1 binomials, taken two and two. . 

In general, since N expresses the sum of the products of the 
second terms of the m binomials, taken m and n, and M the sum 
of their products, taken n—1 and n—1; if we multiply the 
last set by the new second term k, then N+ MA, or the co-effi- 
cient of the term which has n terms before it, will be equal to 
the sum of the different products of the second terms of the 
m-+ 1 binomials, taken n and ». The last term is equal to the 
continued product of the second terms of the m + 1 binomials. 

Therefore, the law of composition, supposed true for a number 
m of binomial factors, is also true for a number denoted by m + 1. 
Hence, it is true for m+ 2, &c., and is therefore general. 


203. Let us now suppose, that in the product resulting from 
the multiplication of the m binomial factors, 


eta ©£t+b, «+c, r+d,.... a 


| 
}CHAP. VIII.] BINOMIAL THEOREM. 203 


ae make: PaaS Oey NT 
-we shall then have 

(a + a) (w+ 6) (@+ec)....-.-.- a (ta). 
‘The co-efficient of the first term, 2”, will become 1. The co-effi- 
cient of a™—1, being a+ 6+c¢-+d,...will be a taken m times; 
that is, ma. The co-efficient of «—*, being 
| ab +ac+.ad.... reduces to at+a?+a@?... 


that is, it becomes a? taken as many times as there are com- 
binations of m letters, taken two and two, and hence reduces 
(Art. 201), to | . 


ah: 1c 

The co-efficient of 2-3 reduces to the product of a%, multiplied 
by the number of different combinations of m letters, taken three 
and three; that is, to 

m—l1 m—2 
| DP / Bh \ AS 
| In general, let us denote the term, which has » terms before 
‘it, by Nav". Then, the co-efficient N will denote the sum of 
‘the products of the second terms, taken ” and x; and when all 
‘of the terms are supposed equal, it becomes equal to a” multiplied 
‘by the number of different combinations that can be made with m 





7 ae, 


m. 








‘letiers, taken n and n. Therefore, the co-efficient of the gineral 
term (Art. 201), is P(m—n+1) 
| ey exp ‘ 
from which we deduce the formula, 
Be 
‘ (2 a)" = 2™+ max™1 +m. — ree 
i 4 
m—1 m—2 pe ar tet) 
pee Aym—3 Ny M—N me 
| SE os x aie Ca a™x Fi PM wn! isd 
The term 
EL) ee 
| Qn 
is called the general term, because by making n= 2, 3, 4, - - 
all of the others can be deduced from it. The term which im- 
mediately precedes it, is evidently, n 
'P 
Q 


r 

— yu"—lym—ntl, since 
- 

: 


204 ELEMENTS OF ALGEBRA. (CHAP. VIM, 
expresses the number of combinations of m letters taken n — 1 
and n—1. Hence, we see, that the co-efficient ~ 
P(m—n+ 1): 
Q Xinel 





ms ae 
is equal to the co-efficient ia) of the preceding term, multiplied 


by m—n-+ 1, the exponent of x in that term, and divided by 
n, the number of terms preceding the required term. 


Since ot is the co-efficient of the preceding term, we may, by 


Q 
observing how the co-efficients are formed from each other, ex- 
press the co-efficient of the general term thus, 


i Mar N/a (m— 3)... (m—n+2)(m—n+ 1) 


The simple law, demonstrated above, enables us to determine 
the co-efficient of any term from the co-efficient of the preceding 
term. 

The co-efficient of any term is formed by multiplying the co-effi- 
cient of the preceding term by the exponent of x in that term, and 
dividing the product by the number of terms which precede the re- 
quired term. 


For example, let it be required to develop (# + a), 
From this law, we have, 


(# + a) = x® + 6ax® + 15a? xt + 20a3x3 + 15atx? + Gada + a’. 


After having formed the first two terms from the general 
formula 2” + max"! +, ... multiply 6, the co-efficient of the 
second term, by 5, the exponent of x in that term, and then 
divide the product by 2, which gives 15 for the co-efficient of 
the third term. ‘To obtain that of the fourth, multiply 15 by 4, 
the exponent of x in the third term, and divide the product by 3, 
the number of terms which precede the fourth; this gives 20; 
and the co-efficients of the other terms are found in the same 
way. 

In like manner, we find 


(w + a)? = x + 10aa® + 450?x8 + 120a3x7 + 210atas 
+ 252a°x° + 210a%a* 4+ 120a7xv3 + 450842 + 10a%x + a9 


‘CHAP. VIII.] BINOMIAL THEOREM. 205 


204. It frequently occurs that the terms of the binomial are 
affected with co-efficients and exponents, as in the following ex- 
‘ample: 

Let it be required to raise the binomial 


3a2c — 2bd 






to the fourth power. 

Placing 3a’c = a2 and —2bd=y, we have 

(x + y)* = at + 4a3y + 6ax?y? + day? + y'; 
‘pe substituting for « and y their values, we have 
Bae —, 2bd)* = (3ac)* + 4 (3a7c)? (— 2bd) + 6 (3a2e)? (— bd)? 
+ 4 (3a%c) (— 2hd)? + (— 2bd)4, 

or, by performing the operations indicated, 

(3a*c — 2bd)* = 8la®ct — 216a%c3bd + 216utc2b2d? — 96a7ch3d 

| + 166d. 

' The terms of the development are alternately plus and minus, 
as they should be, since the second term is —. 
205. The powers of any polynomial, may easily be found by 
‘he binomial theorem. For example, raise 


at+b+e 





© the third power. 
First, put b+c=d. 
Then (a¢+6+c)?= (a+ d)? = a3 + 3a?d + 3ad? + a3; 
ind by substituting for the value of d, 
(a + 6 +c)? = a8 + 30% + 3ab? + 5 
‘Barc + 3b2c + babe 
+ 3ac? + 3c? 
+ ¢3, 


This development is composed of the cubes of the three terms, 


| 


wus three times the square of each term by the first powers of the 
wo others, plus six times the product of all three terms. 

To apply the preceding formula to the development of the cube 
f a trinomial, in which the terms are affected with co-eflicients 
md exponents, designate each term by a single letter, and perform 
he operations indicated; then replace the letters introduced by their 
lalues 





A 


206 ELEMENTS OF ALGEBRA. (CHAP. VIII. 


From this rule, we will find that 
(2a? — 4ab + 36?)3 — 8a — 48a5b + 132atb? — 208233 
+ 198a?b* — 108ab> + 2708. 
The fourth, fifth, d&@c. powers of any polynomial can be developed 
im a similar manner. 


Consequences of the Binomial Formula. 


206. The development of the binomial expression (x + a)” 
will always contain m-+ 1 terms. Hence, if we take that tenn 
of the development which has n terms before it, the number of 
terms after it will be expressed by m— n. 

Let us now seek the co-efficient of the term which has n terms 
after it, and which, consequently, has m—an terms before it. 
We obtain this co-efficient by simply substituting m—n for n, 
in the last value of NV in Art. 203. We then have, 


_m(m—1)(m—2).. (W+2)H$1) 


a 


As we can always take the term which has n terms before it, 
nearer to the first term than the one which has m— n terms be- 
fore it,/we will examine that part of the co-efficient which is 
derived from the terms lying between these two. We may write 
yam): (m—n-+1) .(m—n) .(m—n—]1)...(n+2). 2).(n+B) 

gre. 2 n (n+1) .(m+2) .. (m—n—1) .(m—ny 

Now, by cancelling the like factors in the numerator and de- 
nominator, we have 











In the development of any power of a binomial, the co-efficient s 
at equal distances from the two extremes are equal to each other. 


207. If we der neig o K the co-efficient of the term whieh 
has n terms before it,4that term will be expressed by Ka*a™—"; 
and the corresponding term counted from the last term of the 
series, will be Ka™—*x". 

Now, the first co-efficient expresses the number of different com- 
binations that ,can be formed with 7 ldtters taken n and n; an? 


the second, the number which can be formed when taken m yn 
go* a 










‘HAP. VIII.] CUBE ROOT OF NUMBERS. 207 


md m--n; we may therefore conclude that, the number of dif- 
ferent combinations of m letters taken n and n, is equal to the num- 
rer of combinations of m letters taken m—n and m—n. 

_ For example, ieinelve letters combined 5 and 5, give the same 
vumber of combipations as when taken 12 —5 and 12 — 5, or 7 
and 7. Five letters combined 2 and 2, give the same number of 
‘ombinations as 2 eyo combined 5 — 2 and 5 — 2, or 3 and 3. 


| 208. If, in the, ee formula, 
—1 
(a + a)” — tn + max™—-! + m — azxm™—2 +, &e., 
We suppose x=1, a= 1, we have, 
m— A m —2 
—— . —-— & 
5 +, We, 
That is, the sum of the co-efficients of all the terms hy the for- 
nula for the binomial, is equal to the mth power of 2. 
' Thus, in the particular case 
| (x + a) = a° + 5ax* + 10a%x3 + 10a%x? + Sate +.a°, 
whe sum of the co-efficients 
. 1+5+10+10+5+1 
s equal to 25 = 32. In the 10th power developed, the sum of the 
‘30-efficients is equal to 2!° = 1024. 


) 
(@+ap or ma limtm@— 4m 





Extraction of the Cube Root of Numbers. 


| 209. The cube or third power of a number, is the product which 
‘irises from multiplying the number twice by itself. The cube root, 
lr third root of a number is either of three equal factors into 
\which it may be resolved; and hence, to extract the cube root, 
's to seek one of these factors. 

Every number which can be resolved into three equal factors 






that are commensurable with unity, is called a perfect cube; and 
any number which cannot be so resolved, is called an wmperfect 
| cube. 

' The first ten numbers are 

STS D8. i)4, 5; 6, 7, 8.4 9,) 10} 
Houbes, 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000. 

| Reciprocally, the numbers of the first line are the cube roots 
jof the numbers of the second 















208 ELEMENTS OF ALGEBRA. [CHAP. VIII 


We perceive, by inspection, that there are but nine perfect cube; 
among all the numbers expressed by one, two, and three figures 
Every other number, except the nine written above, which can be 
expressed by one, two, or three figures, will be an imperfect cube 
and hence, its cube root will be expressed by ‘a whole number 
plus an irrational number, as may be shown by a course of rea 
soning entirely similar to that pursued in the latter part of Art. 118 


210. Let us find the difference between the cubes of two con: 
secutive numbers. 

Let a and a+ 1, be two consecutive whole numbers; we have 

(a+ 1)? = a? + 3a2+ 3a+4+1; 

whence, (a+ 1)3 — a = 3a?+ 3a+1.0 


That is, the difference between the cubes of two consecutive whol: 
numbers, 1s equal to three times the square of the least number, plu: 
three times the number, plus 1. 


Thus, the difference between the cube of 90 and the cube of 
89, is equal to ) 
3 (89)? + 3 x 89 + 1 = 24031. 

211. In order to extract the cube root of an entire number, we 
will observe, that when the figures expressing the number do no 
exceed three, the entire part of the root is found by comparin: 
the number with the first nine perfect cubes. For example, th 
cube root of 27 is 3. The cube root of 30 is 3, plus an irration 
number, less than unity. The cube root of 72 is 4, plus an it 
rational number less than unity, since 72 lies between the perfec 
cubes 64 and 125. 

When the number is expressed by more than three figures, the 
process will be as follows. Let the proposed number be 103823 


103 823 | 47 48 Av 

64 3 48 47 

42 x 3 = 48] 398.23 | 384 329 
192 188 

2304 2209 

48 47 

18432 15463 
9216 8836 


110592 103823 
This number being comprised between 1,000, which is the cub 







|OHAP. VIII.) CUBE ROOT OF NUMBERS. 209 


lof 10, and 1,000,000, which is the cube of 100, its root will be 
oxpressed by two figures, or by tens and units. Denoting the tens 
xy a, and the units by 6, we have (Art. 198), 


\ : (a ot b)3 == a? ole 3a2b ote 3ab? + 53, 


Whence it follows, that the cube of a number composed of tens 
and units, is made up of four distinct parts: viz., the cube of the 
‘ens, three times the product of the square of the tens by the units, 
hree times the product of the tens by the square of the units, and 
he cube of the units. 

. Now, the cube of the tens, giving at least, thousands, the last 
i: figures to the right cannot form a part of it: the cube of 
ens must therefore be found in the part 103 which is separated 
‘rom the last three figures. The root of the greatest cube con- 
ained in 103 being 4, this is the number of tens in the required 
oot. Indeed, 103823 is evidently comprised between (40)3 or 
134,000, and (50)? or 125,000; hence, the required root is com- 
yosed of 4 tens, plus a certain number of units less than ¢en. 
| Having found the number of tens, subtract its cube, 64, from 
|.03, and there remains 39, to which bring down the part 823, 
md we have 39823, which contains three times the square of the 
ens by the units, plus the two parts named above. Now, as the 
“quare of tens gives at least hundreds, it follows that three times 
he square of the tens by the units, must be found in the part 
198, to the left of 23, which is separated from it by a point. 
‘Therefore, dividing 398 by 48, which is three times the square 
'f the tens, the quotient 8 will be the units of the root, or some- 
hing greater, since 398 hundreds is composed of three times the 
‘Quare of the tens by the units, together with the two other parts. 

We may ascertain whether the figure 8 is too great, by forming 
rom the 4 tens and 8 units the three parts which enter into 39823; 
/ut it is much easier to cube 48, as has been done in the above ta- 
le. Now, the cube of 48 is 110592, which is greater than 103823; 
aerefore, 8 is too great. By cubing 47 we obtain 103823; hence 
ae proposed number is a perfect cube, and 47 is its cube root. 





Remark |.—The units figures could not be first obtained, because 
ae cube of the units might give tens, and even hundreds, and the 
ms and hundreds would be confounded with those which arise ‘ 


tom other pasts of the cube. 
14 


210 ELEMENTS OF ALGEBRA. [CHAP. VIII, | 


Remark I].—The operations in the last example have been pet- 
formed on but two periods. It is plain, however, that the same| 
reasoning is equally applicable to larger numbers; for, by chan-| 
ging the order of the units, we do not change the relation in 
which they stand to each other. 


plicable to larger numbers. 


212. Hence, for the extraction of the cube root of numbers, we. 
have the following 


RULE. 


I. Separate the given number into periods of three figures each 
beginning at the right hand: the left-hand period will often contain | 
less than three places of figures. | 

Il. Seek the greatest cube in the first period, at the left, and set| 
wts root on the right, after the manner of a quotient in division. Sub- 
tract the cube of this figure of the root from the first period, and to 
the remainder bring down the first figure of the next period, and call 
this number the dividend. 

III. Take three times the square of the root just found for a di- 
visor, and see how often it is contained in the dividend, and place 
the quotient for a second figure of the root. Then cube the figures 
of the root thus found, and if their cube be greater than the first 
two periods of the given number, diminish the last figure ; but of at 
be less, subtract it from the first two periods, and to the remainder 
bring down the first figure of the next period, for a new dividend. 






IV. Take’ three times the square of the whole root for a new divi- 
sor, and seek how often it is contained in the new dividend; the 
quotient will be the third figure of the root. Cube the whole root, 
and subtract the result from the first three periods of the given num | 
her, and proceed in a similar way for all the periods. 


Remark.—If any of the remainders: are equal to, or exceed, 
three tumes the square of the root obtained plus three times this root, 
plus one, the last figure of the’ root is too small and must be aug- 
mented by at least unity (Art. 210). 


AAP. VIII.) EXTRACTION OF ROOTS. 211 


ibe: | EXAMPLES. 
a. 3/48228544. = 364. 

| 2. 4/27054036008 = = 3002. 

| 3. mussgag° = 78, with a remainder 8697. 

4 4/ 91632508641 = 4508, with a remainder 20644129. 
5. 4/32977340218432 = 32068. 


To extract the n* Root of a whole Number. 




















| 213. The n” root of a number, is one of the n equal factors 
jato which the number may be resolved. If the factors are com 
‘ensurable with unity, the number is said to be a perfect power, 
‘they are not commensurable with unity, the number is said to 
je an imperfect power. 

In order to generalize the process for the extraction of roots, 
ye will denote the proposed number by JN, and the degree of the 
pot to be extracted by n. If the number of figures in N, does 
ot exceed n, the rational part of the root will be expressed by 
. single figure. 

Having formed the n” power of all the numbers from 1 to 9, 
iclusive, compare the given number with these powers, and the 
jot of the one next less, will be that part of the required root 
(‘hich can be expressed by a whole number; for, the n” power 
i! 9 is the largest number which can be expressed by » figures. 
When JN contains more than n figures, there will be more than 
hae figure in the root, which may then be considered as composed 
h tens and units. Designating the tens by a, and the units by 4, 
kb have (Art. 203), 


N = (a + b)" = a® + na®—1b + nt a™—2h2 +, &ec.; 


j1at is, the proposed number contains the n® power of the tens, 
us n times the product of the n —1™ power of the tens by the 
juts, plus a series of other parts which it is not necessary to 
)onsider. 

Now, as the n” power of the tens, cannot give units of an or- 
y2r inferior to 1 followed by 2 ciphers, the last m figures on the 
ight, cannot make a part of it. They mustethen be pointed off 


212 ELEMENTS OF ALGEBRA. [CHAP. VII. 


and the root of the greatest power contained in the figures on 
the left should be extracted: this root will be the tens of the re- 
quired root. 

If this part on the left should contain more than n figures, the 
n figures on the right of it, must be separated from the rest, and 
the root of the, greatest n”* power contained in the part on the left 
extracted, and so on. Hence the following 


RULE. 


I. Divide the number N into periods of n figures each, beginning 
at the right hand; eatract the root of the greatest n> power con- 
tained in the left-hand period, and subtract the n™ power of this 
figure from the left-hand period. 

Il. Bring down to the right of the remainder derived from the 
left-hand period, the first figure of the neat period, and call ths 
number the dividend. 

Ill. Form the n —1 power of the first figure of the root, mul- 
tiply it by n, and see how often the product is contained in the divi- 
dend: the quotient will be the second figure of the root, or something 
greater. 

IV. Raise the number thus formed to the n' power, then subtract 
this result from the two left-hand periods, and to the new remainder 
bring down the first figure of the next period: then divide the num- 
ber thus formed by n times the n—1 power of the two figures of 
the root already found, and continue this operation until all the pe- 
riods are brought down. 





EXAMPLES. 
1. What is the fourth root of 531441? 
631441 | 27 
reer acee 16 
4 x 23 = 32 | 371 
(27)4 = 531441. 
We first divide off, from the right hand, the period of four figures 


and then find the greatest fourth root contained in 53, the firs 
period to the left, which is 2. We next subtract the 4th powe 
of 2, which is 16, from 53, and to the remainder 37 we brin; 
down the first figure,of the next period. We then divide 371 by 


f 
} 


| ‘f 


(ai Vult.] EXTRACTION OF ROOTS. 213 











\f times the cube of 2, which gives 11 for a quotient: but. this 
we know is too large. By trying the numbers 9 and 8, we find 
hem also too large: then trying 7, we find the exact root to be 27. 


_ 214. Remarx.—When the degree of the root to be extracted is 
| 1 multiple of two or more numbers, as 4, 6, , the root can be 


obtained by extracting the roots of more ane ee successively. 
- explain this, we will remark that, 


(a3)* = @ X @ X @ X a3 = g3454343 — g3X4 — gl? 
‘nd that in general GArt#'13); 
(a™)" = a™ X a™ X a™ & a™ ©. == mn, 


tence, the n‘* power of the m* power of a number, is equal to the 
lant? power of this number. 


‘ 
} Let us see if the reciprocal of this is also true. 
/ n 

\. Let \/"/a=a; 

{ 
hen raising both members to the ni power, we have, from the 


he 
"la =a’; 


ies tas)? — g/m 
| Extracting the mn’ root of the last equation, we have 


mn ae 
Jena 


and hence, ‘ / ie: pat are 


| ince each is equal to a’. Therefore, the n‘ root of the m*” root 
Ve any number, is equal to the mn root of that number. And in 
} similar manner, it might be proved that 


n / mn / 
a= a. 


' By this method we find that 


pi. V 256 =4/ 4/256 = ./16 = 4. 


| ae a se ERIN, Se ake 
\ 2 “/ 2985984 = 4/./ 2985984 — 3/1728 = 12. 


m 


214 ELEMENTS OF ALGEBRA. [CHAP. VIIl 


leas 3 ie 
3 §/1771561 =4/ + 1771561 = 11. 


4 Y 1679616 = 4/1296 = 1/1296 = 6. 


Remarx.—Although the successive roots may be extracted i1 
any order whatever, it is better to extract the roots of the lowes 
degree first, for then the extraction of the roots of the highe 
degrees, which is a more complicated operation, is effected upo 
numbers containing fewer figures than the proposed number. 


Extraction of Roots by Approximation. 


215. When it is required to extract the n™” root of a numbe 
which is not a perfect power, the method already explained, wi 
give only the entire part of the root, or the root to within unity 
As to the number which is to be added, in order to complete th 
root, it cannot be obtained exactly, but we can approximate to 
as near as we please. 

Let it be required to extract the n” root of the whole numb¢ 


oe ; 1 
a to within a fraction —; that is, so near, that the error sha 


be less than hy 
P 


We will observe that we can write 
ap” 


C= + 
n 





If we denote by 7, the root of ap” to within unity, the numb 


n ” Ly 
a Mas will be comprehended between i and rh 


n n 


therefore the 4/ a will be comprised between the two number 


"and i ale and consequently, their difference ad will | 
p 


r 
greater than the difference between — and the true root. Hence 
P 


ata Bak ; 
— is the required root to within the fraction —. 
? 


Hence, to extract the n root of a whole number to within a fra 
eon 
tion —, multiply the number by p"; extract the n™ root of ¢ 


product to within unity, and divide the result by p. 


CHAP VII.) EXTRACTION OF ROOTS. 215 
216. Again, suppose it is required to extract the n’ root of the 

_ i 

fraction + 


Multiply each term of the fraction by 





: a ab™—1 
b*=!, and it becomes — = —— 
b bn 
Let r denote the n™ root of ab"™—!, to within unity ; 
yo" a rn (r+1)" 








= —, will be comprised between an and 


pn b bn ? 


and consequently, a will be the n™” root of > to within the 


fraction ‘ : 
fraction’ —. 
b 


i Therefore, after having made the denominator of the fraction a 
perfect power of the n‘* degree, extract the n root of the numerator, 
to within unity, and divide the result by the root of the new de- 
nominator. 
When a greater degree of exactness is required than that in- 


ee 1 ih. 
dicated by + extract the root of ab*—! to within any fraction 


] 7 ee 
—; and designate this root by —. Now, since — is the root 
P B 


/ 


yee my Ae refi 
‘of the numerator to within —, it follows, that reas the true 
P 2 


root of the fraction to within be 
Pp 


EXAMPLES. 
1. Suppose it were required to extract the cube root of 15, to 


ae, 1 
within H We have 


orn te do xX 1728 = 20920. 


Now the cube root of 25920, to within unity, is 29; hence 
‘the required root is, 


paerear Ld 
' 2. Extract the cube root of 47, to within — 


20° 
We have 
47 x 203 = 47 x 8000 = 376000. 


216 ELEMENTS OF ALGEBRA. (CHAP. VII, 


Now the cube root of 376000, to within unity, is 72; hence 
12 12 “OE 1 
/ 4 = rhe 30 to within 50° 
3. Find the value of 1/25 to within 0.001. 


To do this, multiply 25 by the cube of 1000, or 1000000000, 
which gives 25000000000. Now, the cube root of this number, 
is 2920; hence 


»/ 25 = 2.920 to within 0.001. 


217. Remarx.—lIn general, in order to extract the cube root of a 
whole number to within a given decimal fraction, annex three times as 
many ciphers to the number, as there are decimal places in the required 
root; extract the cube root of the number thus formed to within 


unity, and point off from the right of this root the required number 
of decimals. 


218. We will now explain the method of extracting the cube 
root of a decimal fraction. Suppose it is required to extract the 
cube -root of 3.1415. 

Since the denominator, 10000, of this fraction, is not a per- 
fect cube, make it one, by multiplying it by 100; this is equivalent 
to annexing two ciphers to the proposed decimal, which then be- 
comes, 3.141500. Extract the cube root of 3141500, that is, of 
the number considered independent of the comma, to within unity; 


this gives 146. Then divide by 100, or +/ 1000000, and we find 
‘/ 3.1415 = 1.46 to within 0.01. 


Hence, to extract the cube root of a decimal number, we have 
the following 


RULE. 

Annex ciphers to the decimal part, if necessary, until at can be 
divided into exact periods of three figures each, observing that the 
number of periods must be made equal to the number of decimal 
places required in the root. Then, extract the root as tn entire num- 
bers, and point off as many places for decimals as there are periods 
in the decimal part of the number. 

To extract the cube root of a vulgar fraction to within a given 
decimal fraction, the most simple method is to reduce the pro- 
posed fraction to a decimal fraction, continuing the operation until 
















HAP. VIII.] EXTRACTION OF ROOTS. 217 


\he number of decimal places is equal to three times the number re- 
|uired tn the root. ‘The question is then reduced to extracting the 
ube root of a decimal fraction. 

| 219. Suppose it is required to find the sixth root of 23, to 
\ithin 0.01. 

| Applying the rule of Art. 215 to this example, we multiply 23 
y 1008, or annex twelve ciphers to 23; then extract the sixth root 
) the number thus formed to within unity, and divide this root 
y 100, or point off two decimals on the right. 


We thus find that 6/23 = 1.68, to within 0.01. 


EXAMPLES. 
1. Find the */473 to within J. Ans. 73. 
2. Find the {/79 to within .0001. Ans. 4.2908. 
3. Find the 4/13 to within .01. Ans. 1.53. 
M4. Find the 4/3.00415 to within .0001. Ans. 1.4429. 
5. Find the 9/0.00101 to within .01. Ans. 0.10. 
6. Find the / itt: to within .001. Ans. 0.824. 


Extraction of Roots of Algebraic Quantities. 


) 220. Before extracting the root of an algebraic quantity, let us 
}e¢ in what manner any power of it may be formed. 

Let it be required to form the fifth power of 2a342, We have 
(2a%b?)> — 2a%b? x 2ab? x 2a3b? x 2a3h2 x 2a3h2, 

’m which it follows, 1st. That the co-efficient 2 must be mul- 
| lied by itself four times, or raised to the 5th power. 2d. That 


| : 
ich of the exponents of the letters must be added to itself four 


| Hence, (297) —= 25., a3X5h2X5 — 39q15510, 

hIn like manner,  (802b3c)3 = 83 , a2X3)3X3¢3 — 512q5%3, 
‘Therefore, in order to raise a monomial to any power, raise 
}' co-efficient to this power, and multiply the exponent of each of 
| letters by the exponent of the power, and unite the terms. 
}Hence, to extract any root of a monomial, 

hIst. Extract the root of the co-efficient and divide the exponent 
feach letter by the index of the root. 2d. To the root of the co- 


218 ELEMENTS OF ALGEBRA. [CHAP. VIII. 


efficient annex each letter with its new exponent, and the result wil 


be the required root. ‘Thus, , 


5/ 640988 = 4a%bc? ; “/ 16a8b%ct = 2abc. 


From this rule we perceive, that in order that a monomial maj 
be a perfect power, lst, its co-efficient must be a perfect power 
and 2d, the exponent of each letter must be divisible by the a 
dex of the root to be extracted. It will be shown hereafter, hov 
the expression for the root of a quantity, which is not a perfec 
power, is reduced to its simplest terms. 


221. Hitherto, in finding the power of a monomial, we hav 
paid no attention to the sign with which the monomial may b 
affected. It has already been shown, that whatever be the sig 
of a monomial, its square is always positive. 

Let » be any whole number; then, every power of an eve 
degree, as 2n, can be considered as the n” power of the square 
thatiis,s\(a? tea". 

Hence, it follows, that every power of an even degree, will be e: 
sentially positive, whether the quantity itself be positwe or negative 


Thus, (+ 2a2b%c)t = + 16a%b'%ct. 


Again, as every power of an uneven degree, 2n + 1, is but th 
product of the power of an even degree, 2n, by the first powel 
it follows that, every power of an uneven degree, of a monomial, | 
affected with the same sign as the monomial itself. 


Hence, (+ 4a7b)? = + 640°? ; and a 4a*b)3 = — 64a%53 
From the preceding reasonings, we conclude, ; 


Ist. That when the degree of the root of a monomial is uneve 
the root will be affected with the same sign as the monomial. 


Hence, 


V+ 8a—+2a; /— 8a = —2a; 4/— 32u%5 = — 20%, 


2d. When the degree of the root is even, and the monomial a po: 
tive quantity, the root is affected either with the sign + or —. 


Thus, 4/ 81th? = +3ab3; 4/64a!8 = + 2a3. 


3d. When the degree of the root is even, and the monomial ney 
tive, the root is impossible; for, there is no quantity which, beit 


‘CHAP. VIII.] EXTRACTION OF ROOTS. 219 


‘raised to a power of an even degree, will give a negative result. 


Therefore, 
Gamers = Fi 8h 


,ace symbols of operation which it is impossible to execute. They 


are wunaginary expressions (Art. 126), like . 


| ae 2 me 











EXAMPLES. 
1. What is the cube root of 8a%b3c!?? Ans. 2abec*. 
2 What is the 4th root of &la*tb®c!6? Ans. 3ab?e‘. 
3 What is the 5th root of — 32a5cl%q15? Ans. — 2ac?d3. 
4. What is the cube root of — 125a9b%c3?  =Ans. — 5a3b?e. 


Extraction of Roots of Polynomaals. 

222. Let us first examine the law of formation of any power 
‘of a polynomial. To begin with a simple example, let us develop 
(a + y + 2). 

If we place y+ 2-= 1, we shall have, 
(a + u)> = a + 3a7u + 3au? + v3; 
or by replacing u by its value, y + 2, 
(at y +2) =a + 3a? (y + 2) + 3aly + 2)? + (y + 2)5 


| or performing the operations indicated, 
: 
L 





(a +y+zp= a? + 3a°y + 3a?z + 3ay? + 6ayz + 3az? + yP+ 3y2z2 


+ 3yz2? -+ 2°. 
When the polynomial is composed of more than three terms, as 
jat+yt2ztnr....p, let, as before, u = the sum of all the 





terms after the first. Then, a+ u will be equal to the given 

polynomial, and 
(a + u)? = a3 + 8a?u + 3au? + U3; 

from which we see, that the cube of any polynomial is equal to the 

cube of the first term, plus three times the square of the first term 

multiplied by each of the remaining terms, plus other terms. 

If « does not contain a, it is plain that the exponent of @ in 
each term, as a?, 3a2u, d&c., will be greater than in any of the 
following terms; and hence, every term will be irreducible with the 
‘terms which precede or follow it. 





{ 


220 ELEMENTS OF ALGEBRA. [CHAP. VII. 


If wu contains a, as in the polynomial 
a®*+acr+ 6, where u=ar+Q8, 


the terms will still be irreducible with each other, provided we 
arrange the polynomial with reference to the letter a. For, if the 
given polynomfal be arranged with reference to a, the exponen’ 
of a in the first term will be greater than the exponent of a ix 
w: hence, its cube will contain @ with a greater exponent thar 
will result from multiplying its square by u. Also, the co-efficien 
of w multiplied by the first term of uw, will contain a to a higher 
power than any of the following terms of the development, anc 
hence, will be irreducible with them; and the same may be showr 
for the subsequent terms. | 

In order to extract any root of a polynomial, we will first ex 
plain the method of extracting the cube root. It will then be 
easy to generalize this method, and apply it to the case of any 
root whatever. 

Let WN be any polynomial, and R its cube root. Suppose the 
two polynomials to be arranged with reference to some letter, as 
a, for example. It results from the law of formation of the cube 
of a polynomial (Art. 222), that in the cube of R, the cube of 
the first term, and three times the square of the first term by the 
second, cannot be reduced with each other, nor with any of the 
following terms. 

Hence, the cube root of that term of N which contains a, af 
fected with the highest exponent, will be the first term of R; anc 
the second term of R will be found by dividing the second tern 
of IV by three times the square of the first term of R. 

By examining the development of the trinomial @ + y+ 2, we 
see, that if we form the cube of the two terms of the root founc 
as above, and subtract it from NV, and then divide the first tern 
of the remainder by 3 times the square of the first term of R 
the quotient will be the third term of the root. Therefore, having 
arranged the terms of N, with reference to any letter, we have 
for the extraction of the cube root, the following 


RULE. 


1. Extract the cube rovt of the first term. 
Il. Divide the second term of N by three times the square of th 
first term of R; the quotient will be the second term of R. . 


| 
‘HAP. VIII.] EXTRACTION OF ROOTS. S24 


ML. Having found the first two terms of R, form the cube of this 
imomal and subtract it from N ; after which, divide the first term 
f the remainder, by three times the square of the first term of R- 
‘he quotient will be the third term of R. 

| IV. Cube the three terms of the root found, and subtract the cube 
‘rom N: then divide the first term of the remainder by the divisor 
‘lready used, and the quotient will be the fourth term of the root: 


‘he remaining terms, if there are any, may be found in a similar 
wanner. 
i 

’ EXAMPLES. 

1. Extract the cube root of «°— 6x5+152t— 2023+. 15”2—6x-+1. 
x® — 629+ 1524 — 2023 + 1542 —6r+1 |x? — Qe + 1 
 (# — 22)? = 18 605 4+ 1221— 823 Sree BVT” 
‘st rem. me eT eg Ta 1243+, &e. 

P—2x2+1)? = 2° — 6x5 + 150+ — 2003 + 1502— 62 +1. 








| In this example, we first extract the cube root of x6, which 
ives x”, for the first term of the root. Squaring x?, and multi- 
lying by 3, we obtain the divisor 3x+: this is contained in the 
econd term — 6x5, — 2x times. Then cubing the root, and sub- 
acting, we find that the first term of the remainder 3x‘, contains 
ie divisor once. Cubing the whole root, we find the cube equal 
) the given polynomial. Hence, x?— 2x +1, is the exact cube 
0b. 
| 2. Find the cube root of 

x6 + 625 — 40x? + 96a — 64. 
‘3. Find the cube root of 
a 8a® — 12x54 3024 — 25”3 + 302? — 12 + 8 


223. The rule for the extraction of the cube root is easily ex- 
nded to a root with a higher index. For, 


Let Gree c+ 2. f, be any polynomial. 
Let s= the sum of all the terms after the first. 
Then a+ s = the given polynomial; and 
(a+ s)" =a" + na"—15 + other terms. 
That is, the n™ power of a polynomial, is equal to the un power 
‘the first term, plus n times the first term raised to the power 


: 


222 ELEMENTS OF ALGEBRA. |CHAP. VIM. 


n—1, multiplied by each of the remaining terms, + other terms 
of the development. 

Hence, we see, that the rule for the cube root will become the 
rule for the n” root, by first extracting the n™” root of the first 
term, taking for a divisor 2 times this root raised to the n— 1 
power, and raising the partial roots to the n™ power, instead of 
to the cube. 


EXAMPLES. 


1. Extract the 4th root of 
16a* — 96a%x + 216a2x? — 216ax? + 81zx4. 
16a*— 96a3x-+ 216a2x?—216ax?+ 8lat | 2a—3ax 


(2a — 3x)* = 16a*— 96a3x+-216a?x? —216ax3+ 81at | 4 x (2a)? =32a? 


We first extract the 4th root of 16a*, which is 2a. We then 
raise 2a to the third power, and multiply by 4, the index of the 
root; this gives the divisor 32a3. This divisor is contained in 
the second term — 96a3x, —- 3x times, which is the second term 
of the root. Raising the whole root to the 4th power, we find 
the power equal to the given polynomial. 


2. What is the 4th root of the polynomial, 
8la%ct + 16h4d! — 96a%eb3d3 — 216a%e3bd + 216atc?b?d?. 


3. Find the 5th root of 
32x25 — 80xt + 80x23? — 40x? + 102 — 1. 


Calculus of Radiata 


224. When the monomial or polynomial whose root is to be ex- 
tracted, cannot be resolved into as many equal and rational fac- 
tors as there are units in the index of the root, it is said to be 
an imperfect power. ‘The root is then indicated by placing the 
quantity under the radical sign, and writing over it at the left 
hand, the index of the root. Thus, the fourth root of 3ab? + Qac’, 


is written 
V/ 3ab? + Yac5, 


The index of the root is also called the index of the radical. 
It is plain that a monomial will be a perfect power, when the 
numerical co-efficient is a perfect power, and the exponent of each 
letter exactly divisible by the index of the root. 









HAP. VII.) CALCULUS OF RADICALS. 223 


; By the definition of a root (Art. 213), we have 


) (‘Vabe... 0 gh a tS 


id by the rule for the raising of powers, 


Jax'bx Ve...n=(‘V a)" x (9/ 5)" x (/ c)*... =abe...; 


id since the x” powers are equal, the quantities themselves are 
jual: hence, 


Wabews..aVaxVbxVe... 


at is, the n‘* root of the product of any number of factors, is equal 
the product of their n™ roots. 


1. Let us apply the above principle in reducing to its sim 


“est form the imperfect power, V 54a*b3c?. We have 


‘/ 54atb3e? = — V/ 270363 x ‘/ 2ac? = = 3ab ‘/ 2ac?. ac*. 


|. 2. In like manner, 

‘/ 8a? aie V a? ; and 4/48a58c° — 2abc 4/3ac? ; 
i 3. Also, 
§/192a7be!? — “/ 64a%e!2 x §/3ab = 2ac* ‘/3ab. 

| In the expressions, 3ab ‘/ 2ac?, 2 ‘/ a2, 2Qab2c 4/3ac?, each quan- 
ity placed before the radical, is called a co-efficient of the radical. 


225. The rule of Art. 214 gives rise to another kind of sim- 
‘ification. 


| Take, for example, the radical expression, 9/ 4a?; from this 


tle, we have 
3 
Vv 4? = RY 4a’, 


jd as the quantity affected with the radical of the second de- 
‘ee, “/3 is a perfect square, its root can be extracted : hence, 


Vv 4a? = \/ 2a. 
| In like manner, 
4/3602? —// 36020? = »/6ab. 











\ In generAl, 


™ 
mn m 
/ (1 hae “/ = vV a’ 














224 ELEMENTS OF ALGEBRA. (CHAP. Viti 


that is, when the index of a radical is a multiple of any numbe 
n, and the quantity under the radical sign is an exact n” power 
we can, without changing the value of the radical, divide its inde 
by n, and extract the n” root of the quantity under the sign. 

This proposition is the inverse of another, not less important 
viz., the index of a radical may be multiplied by any number, pro 
vided we raise the quantity under the sign to a power of which thi 
number is the exponent. 

For, since a is the same thing as / aX, we have, 


fa Oa a 


226. This last principle serves to reduce two or more radical 
to a common index. 

For example, let it be required to reduce the two radicals 

V2a and 4/(a+ b) 

to the same index. 

By multiplying the index of the first by 4, the index of the sec 
ond, and raising the quantity 2a to the fourth power; then multi 
plying the index of the second by 3, the index of the first, anc 


cubing a+ b, the value of neither radical will be changed, anc 
the expressions will become 


4/20 re 1 tat = ny 16a*; and V (a + b) = ™/ (a + by3, 


Hence, to reduce radicals to a common index, we have the fol 
lowing | 


RULE. 


Multiply the index of each radical by the product of the indier. 
of all the other radicals, and raise the quantity under ‘each radice 
sign to a power denoted by this product. | 


This rule, which is analogous to that given for the reductior 
of fractions to a common denominator, is susceptible of simila 
modifications. 

For example, reduce the radicals 

Va, Yo, VEER, 
to the same index. ae 1 

Since 24 is the least common multiple of the indices 4, 6, an 

8, it is only necessary to multiply the first by 6, the second by 


;| 
t 


‘HAP. VIII.] CALCULUS OF RADICALS. 225 


, and the third by 3, and to raise the quantities under each rad- 
sal sign to the 6th, 4th, and 3d powers respectively, which 
‘ives 


Va "Va; VI Vo, VF FH = VF PY, 
! 


In applying the above rules to numerical examples, beginners 
ery often make mistakes similar to the following: viz., in redu- 
‘ing the radicals 2 and 1/3 to a common index, after having 
wltiplied the index of the first, by that of the second, and the 
idex of the second by that of the first, then, instead of multiply- 
1g the exponent of the quantity under the first sign by 2, and the 
tponent of that under the second by 3, they often multiply the 
jwantity under the first sign by 2, and the quantity under the sec- 
ad by 3. Thus, they would have 


d=? x2=5/4, and /3 = 9/3 x%3=5/9. 
Whereas, they should have, by the foregoing rule, 
I V2=V@) =Vs md (Sa5/ OF HV. 
Reduce V2, V4, V4, to the same index. 





Addition and Subtraction of Radicals. 


227. Two radicals are similar, when they have the same index, 
‘ad the same quantity under the sign. ‘Thus, 


3+~~ab and 7+/ab; as also, 3a? V 62, and 9c? / b2, 


‘ce similar radicals. 
‘In order to add or subtract similar radicals, add or subtract their 
efficients, and to the sum or difference annex the common radical. 


Thus, 
CASS 3 eee 54/ b; also, ney Aes ed AN ARS 
Again, 3a / b+ 2c Vb = (3a + 2c) Oh 


Dissimilar radicals may sometimes be reduced to similar radi- 
ils, by the rules of Arts. 224 and 225. For example, , 


1 /48ah? + bf 750 = 4b 30 + 5h 3a = 9b /3a 
| [15} 








226 ELEMENTS OF ALGEBRA. [CHAP. VIf. 


2. 4/80% + 16at — 4/o* + 2083 = 2a9/b + 2a — bb + 205 
= (2a — b) /b + 2a. 
3 / 4a? + 24/20 = 39/204 25/20 = 5 V/2a. 


When the radicals are dissimilar and irreducible, they can only 
be added or subtracted, by means of the signs + or —. 


Multiplication and Division. 


228. We will suppose that the radicals have been reduced to a 
common index. 


Let it be required to multiply 4/a@ by ‘V/ 8. . 
If we denote the product by P, we have 


AEN, */ bi 


and by raising both members to the n power, 


(Via) x (/b)" = ab = Pa; 


and by extracting the .n‘ root, 


ax b= P= Va; 


that is, the product of the un roots of two quantities, is equal to the 
n™ root of their product. 


Let it be required to divide “Ya by ‘V/ 5. 
If we designate the quotient by Q, we have 


’ . —=Q; 

n/ b | 
and by raising both members to the n™ power, 
Cai Wigs 
aid six ait fh ay ) 
(Vb)" 6 


and by extracting the n™ root, 


ona | fal 
vee 


that is, the quotient of the n™ roots of two quantities, 1s equal to 
the n‘ root of their quotient. 











‘HAP. VIII.) CALCULUS OF RADICALS. 297 


| Therefore, for the multiplication and division of radicals, we 
ave the following 
RULE. 


I. Reduce the radicals to a common index. 


Il. If the radicals have co-efficients, first multiply or divide them 
eparately. 

III. Multiply or divide the quantities under the radical sign by 
wh other, and prefix to the product or quotient, the common radi- 


a sign. 


EXAMPLES. 


1. The product of 
; 3 /q2 + 22 a ae (a? + ey 2 on (a? + 2) 
c d cd 


8) 6a? (a? + 5?) 
=— coterie : 
2. The product of 
3a 4/ 8a? x 2b 9/ 4a%c = 6ab “/ 32atc = 12a%b V 2c. 


3. The quotient of 
V ab? + bt / 8b (a2? + Bt) 7) Gi + B 


alk we a — ots a? — $2 
=a. 
4. The product of 


3a$/b x 5b5/ 2c = 15ab x *4/8b4c3. 
4 3 
5. Multiply 2 x 7/3 by Af x sf 








Ans. oy 
Ans. 6 5/ 337500. 


6 Multiply 24/15 by 34/10. 


5 
7. Multiply 1/2 by 2/3. 


; Ans. 8 


10 27 
256 


228 ELEMENTS OF ALGEBRA. (CHAP. Viil 


2f3xV4 











8. Reduce ——== += to its lowest terms. 
V2 x3 | 
‘Ans. 412/288 
3 
9. Reduce aes : v3 to its lowest terms. 
f 2x/3 
“ang DA) 2 
sid 3 


10. Multiply af 2, ay 5° and 4/S' together. 


Ans. '4/648000. 


7 3 
4 1 re 
11. Multiply ve a and “/ 6, together. 
42 
Ans, 2 
27 


4 13 
Ans. = te rs oy 42. 


I 1 
13. Divide es = vy (v2+ 34/1) 


14. Divide 1 id Ta + Vt 
Ans. Vai — fat 1 ei +e al 
Tig eae eg 
15. Divide /a+‘%/b by Va—Vb. 
a at+b+2/fab +2 a% +27 ab 


a—b 


Ans. I0" 


Powers and Roots of Radicals. 
229. By raising “/ a to the n” power, we have 


(Vap=VaxVaxVa...=Vam * 


by the rule just given for the multiplication of radicals. Hence, 
for raising a radical to any power, we have the following ) 


ws es Se 
* ty - 

ted 4 . 

fr ) 

, 7 

i a | 
‘e 

s ve | 


‘HAP. VIII.) CALCULUS OF RADICALS. 229 


RULE. 


Raise the quantity under the sign to the given power, and affect 
xe result with the radical sign, having the primitive index. If it 
as a co-efficient, first raise it to the given power. 


EXAMPLES. 


1 Via) = Ve = To = a A 
: (3 9 2a)> = 39. 5/(2a)? = 243 9/3205 — 4860 3/ 402. 
‘V hen the index of the radical is a multiple of the power to 
hich it is to be raised, the result can be simplified. 

Re 9/ oq = / 2a (Art. 214): hence, in order to square 
la we have only to omit the first radical, which gives 


: (2a)? = 2a. 
| Again, to square 4/38, we have / 3b = \/ /3b: hence, 
(9/30)? = 30. 


Consequently, when the index of the radical is divisible by the 
cponent of the power to which it is to be raised, perform the di- 
iston, leaving the quantity under the radical sign unchanged. 


Let it be required to extract the m* root of the radical */ a. 


Ve have (Art. 214), 
s/Va="V/ a. 


Hence, to extract the root of a radical, multiply the index of the 
tdical by the index of the root to be extracted, leaving the quan- 


ty under the sign unchanged. 
This rule is nothing more than the principle of Art. 214, enun- 


ated in an inverse erder. 


A/a 2) oa; and. /3/S0— 8/ G0. 


When the quantity under the radical is a perfect power, of the 
gree of either of the roots to be extracted, the result can be 


mplitied, 








230 ELEMENTS OF ALGEBRA. [CHAP. VII 


Thus, 4/ 8a3 om V 8a? = V 24. 
In like manner, VV 9a? = hav atin V 3a. 


230. The rules just demonstrated for the calculus of radical 
depend upon the fact, that the n root of the product of sever: 
factors, is equal to the product of the n” roots of these factor 
This, however, has been proved on the supposition that, when 7 
powers of the same degree of two expressions are equal, the expre. 
sions themselves are also equal. Now, this last proposition, whic 
is true for absolute numbers, is not always true for algebraic. e: 
pressions ; for it is easily shown that the same number can hay 
more than one square root, cube root, fourth root, Fe. 

Let us denote the algebraic value of the square root of a t 
x, and the arithmetical value of it by p; we have the equations 


a2 —a, and a?=p*, whence x= +p. 


Hence we see, that the square of p, (which is the root of a), w 
give a, whether its sign be + or —. 

In the second place, let x be the algebraic value of the cul 
root of a, and p the numerical value of this root; we have tl 
equations : 

23 —a, and «2° =p’. 

The last equation is satisfied by making x =p. 

Observing that the equation 2? = p> can be put under the for 
x3 — p? = 0, and that the expression x* — p® is divisible by 2 — 
(Art. 61), which gives the exact quotient, x? + px + p*, the abo 
equation can be transformed into 

(x — p) (a? + px + p*) = 0. 

Now, every value of x which will satisfy this equation, w 
satisfy the first equation. But this equation can be satisfied i 
supposing 

2—p=0, whence s>p; 
or by supposing 
| a2 + px + p? = 0, 


from which last, we have 


eee a 


; 
| 
i 


" 
(HAP. VIII.) CALCULUS OF RADICALS. 231 


; Hence, the cube root of a, admits of three different algebraic 
alues, v1z., 


Be) = (4) 





P; 


' Again, resolve the equation 


at = p', | 
i which p denotes the arithmetical value of V a. This equation 
an be put under the form 
i at —pt=0; 
hich reduces to 

(a? — p?) (a? +p?) = 05 
‘nd this equation can be satisfied, by supposing 


: rp —.0, ,whence # = + p; 
r by supposing, 
Zp =, whence # = +f—p=+tpV—1 
We therefore obtain four different algebraic expressions for the 
ourth root of a. 
As another example, resolve the equation 
| x — p® = OU. 


This equation can be put under the form 


| (x3 — ite 7) a 0 


‘hich may be satisfied by making either of the factors equal to 
ero. 








' But | x3 — 3 —0, gives 
: droge lee lining 
fea) gf 9 ( 5 ). 
And if in the equation 23+ p? =0, we make p = — p’, it be- 


‘omes a? — p? = 0, from which we deduce 


adatom, 


2 











, &=p’, and 2 =p'( 


-r, substituting for p’ its value — p, 


adap at ay 


2 


Therefore, the value of # in the equation a® — p*=0, and con- 








o=—p, and 2 = ~ p( 


232 ELEMENTS OF ALGEBRA. [CHAP. VIit 


sequently, the 6th root of a, admits of six values. If we make 


pra ae hy AES — iSaeee ea 


C= 5) 3 and: Ja a 3 ’ 


these values may be expressed by 

















Ps ap, ap, — p — ap — ap. 

We may then conclude from analogy, that in every equation of 
the form a” —a=0, or a” —p™=0, a is susceptible of m dif 
ferent values; that is, the m” root of a number admits of 7 
different algebraic values. 





231. If, in the preceding equations, and the results correspond. 
ing to them, we suppose, as a particular case, a= 1, whence 
p= 1, we shall obtain the second, third, fourth, a&c. roo‘s of 
unity. ‘Thus +1 and —1 are the two square roots of unity 
because the equation #2 —1=—0, gives r= +1. 

In lke manner, 

—l++vy-—3 —-l-—-vy-3 
7; A eS ae ’ 


are the three cube roots of unity, or the roots of 22 —1=0; an 


ues coe a ore — Vea 


are the four fourth roots of unity, or the roots of xt —1= 0, 



















232. It results from the preceding analysis, that the rules fo 
the calculus of radicals, which are exact when applied to abso 
lute numbers, are susceptible of some modifications, when applied t 
expressions or symbols which are purely algebraic; these modifica 
tions are more particularly necessary when applied to imaginar: 
expressions, and are a consequence of what has been said 1 
Art. 230. 

For example, the product of 


J—a by /—a, 


by the rule of Art. 228, would be 


f= 3 Xk 


Now, vy @ is equal to +a (Art. 139); there is, then, ap 
parently, an uncertainty as to the sign with which @ should be 
affected. Nevertheless, the true answer is —a: for, in orde. 








“HAP. VIII.) CALCULUS OF RADICALS. 233 


‘0 square +/ m, it is only necessary to suppress the radical; but 


VY —axvy —a=(¥y —ayPr=—a. 
Again, let it be required to form the product 


a RAY eth 


By the rule of Art. 228, we shall have 


¥—ax¥—b=V+ab. 


Now, /ab= +p (Art. 230), p being the arithmetical value 
f the square root of ab; but the true result is —p or — A/, ab, 
My long as both the radicals nf, —a and ~¥ —b are affected 
vith the sign +. 

For, / —a=V¥V a.f/—1; and + i Oy A ety ES, 

ence, 

\ Al i b= Va. Vv¥—-1xv bx of—1=Vab(f— 1) 
ab x —1=—vab. 


By similar methods we find the different powers of /— 1 to 
e as follows :— 


| 
| 


f 
\ 


ah ¥-1xV-1=(V-1P?=-1. 
2. (f/— 18 =(/—1)?.¥ —l=—-vV—-1. 
| 8. (f/— 1)' =(f— 1). (/— 1? = -—1 x -1=41. 


, Again, let it be proposed to determine the product of cy Ay 


y the +/—b which, from the rule, would be A/ tee ab, and con- 
equently, would give the four values (Art. 231), 


1 og VEE VY EO 


To determine the true product, observe that 

Wiel amghA En, wind S/o 24 out OT. 
ee Gy ~ (\/ / 1) = 
lence 6 OEE pag a 














234 ELEMENTS OF ALGEBRA. (CHAP. VITi. 


We will apply the preceding calculus to the verification of the 


MAS Ear erer ah 
2 Etat 
considered as a root of the equation #3 —1=0; that is, as one 
of the cube roots of 1 (Art. 231). 
. From the formula, 


(a+ b)3 = a® + 3a7b + 3ab? +8°, 


we have we in —_ . 


(=1)8+ 3(—1). V— 3 +.3(— 1) - (= 38) + WW — 3) 
8 


expression 








ti+3 7-3-3. —3 Sy 











= —=1. 
8 8 
—l—vy-—3 
The second value, nae We be verified in the same 


manner. It should be remarked, that either of the imaginary roots 
is the square of the other; a fact which may be easily verified. 


Theory of Exponents. 


233. In extracting the n” root of a quantity a, we have seer 
that when m is a multiple of n, we should divide the exponent a 
by n the index of the root. When m is not divisible by x, the 
operation of extracting the root is indicated by indicating the di 


vision of the two exponents. ‘Thus, 
m 


Va" =a", 
a notation tounded on the rule for the exponents, in the extrac 
tion of the roots of monomials. In such expressions, the numerato 
indicates the power to which the quantity is to be raised, and thi 
denominator, the root’ to be extracted. 


ae 2 4 os 
Therefore, “/ a = a"; and , +/ aes 


If it is required to divide a” by a”, in which m and n are posi 
tive whole numbers, we know that the exponent of the diviso 
should be subtracted from the exponent of the dividend, and we 
have 





HAP. VIII.) THEORY OF EXPONENTS. 235 


‘f m>n, the division will be exact; but when m <a, the di- 
rision cannot be effected, but still we subtract, in the algebraic 
sense, the exponent of the divisor from that of the dividend. Let 
> be the arithmetical difference between n and m; then will 





q™ 
n—=m+j; whence —a-?P: 
+ P; tg ; 
a™ 
yut —— =-—; hence, a? =—. 
qmtp QP aP 


] 

Therefore, the expression a~? is the symbol of a division which 
yas not been performed; and its true value is the quotient repre- 
sented by unity, divided by a, affected with the exponent p, taken 
rositively. ‘Thus, 


| —; and a= —. 


: 1 1 
Since, a-? =—; and -——=aqp?, we conclude that, 
aP ane 
Any factor may be transferred from the numerator to the denom- 
f, é ° 
nator, or from the denominator to the numerator, by changing the 


“ign of its exponent. 


eh. : 1 
If it is required to extract the n™ root of —, we have 
a 


pe eo =. 
—=a-™; hence, mee a/ a Se 

. The notation of fractional exponents, whether positive or nega- 
ive, has the advantage of giving an entire form to all expres- 
slons whose roots or powers are to be indicated. 

From the conventional expressions founded on the preceding rules, 
‘we have 


m n 
me! a t 1 } n 
a=va";, a?=—; and —— = 4." 
“ qP / 


We may therefore substitute the second value in each expres- 


sion, for the first, or reciprocally. 
As a? is called a to the p power, when p is a positive whole 


wmber, so, by analogy, 


i 
y 
| 
: 


m ™ 


a”, a?, a ae, 


e 


° m 
are called, respectively, a to the — power, a to the — p power 
n 


236 ELEMEN'TS OF ALGEBRA. [CHAP. VIII. 


a to the —— _ power, in which algebraists have generalized the 
n 
word power. It would, perhaps, be more accurate to say, a, ex- 
m m 
ponent —, a, exponent —p, and a, exponent — —; using the 
n n 
word power only when we wish to designate the product of a 


number multiplied by itself two or more times. 


Multiplication of Quantities affected with any Exponenis. 


2 gah oi 
234. In order to multiply a® by a%, it is only necessary to add 
the two exponents, and we have 


3 2 
a®’ x a? = 


For, by Art. 233, 
3 2 
a® =/ a3; and a = V arty 
gy | . 
hence, a? xg? = V rt 4 V a2, 


reducing to a common index (Art. 226), and then multiplying, 


342 
aie 


3. 2 a; 12 
; : . na: 5 
Again, multiply a_*_byp a's 
4 
3 1 5 
| -+ ae 
We have, a*= V4, and a® = V5; 


m mm np-— 
p 2 p—mq 


a” xX ai=a" {= @ee 
Therefore, in order to multiply two monomials affected wit! 


any exponents whatever, follow the rule given in Art. 41, for quantitie: 
affected with entire exponents. 


HAP. VIII.) THEORY OF EXPONENTS. 237 


| an this rule, we shall find 


-2 
| ‘a Pe tent x abics — =a a bee 
| : 2 many _14 7 
ies 3a—2b3 x 2a °b2c2 = Ga * hé 2, 
ea] 1 feces 
3. 6a 2b4c—™ x 5ath—-5ce" = 30a Sh-len—m, 


Dinision of Quantities affected with any Exponents. 


235. To divide one monomial by another when both are affected 

vith any exponent whatever, divide the co-efficient of the divi- 
end by that of the divisor, for a new co-efficient: subtract the 
uponent of each letter in the divisor from the exponent of the same 
itter in the dividend, and then annex each letter with its new ex- 
onent. 
For, the exponent of each letter in the quotient must be such, 
iat, added to the exponent of the same letter in the divisor, the 
am shall be equal to the exponent of the letter in the dividend ; 
ence, the exponent of any letter in the quotient, is equal to the 
ifference between the exponent of that letter in the dividend and 
\ the divisor. 


EXAMPLES. 
i. (Bel a ae | aS oF 
2. at = Ohi Be 
3. ab x be = vane = qloz8 
MS Divide, 320788? by 80%55c 2. Ans. iautnee 
| 5. Divide | 64a%2c 8 by 320% Be 8, Ans, 2a'8}5, 


Formation of Powers. 


236. To form the m” power of a monomial, affected with any 
{ponents whatever, raise the co-efficient to the m” power, and 
t the exponents, observe the rule given in Art. 220, viz, multiply 
@ exponent of each letter by the exponent m of the power. 

For, to raise a quantity to the m” power, is the same thing as 
' multiply it by itself m—1 times; therefore, by the rule for 
‘ultiplication, the exponent of each letter must be added to itself 
“= 1 times, or multiplied by m. 


238 ELEMENTS OF ALGEBRA. (CHAP. VIII. 


3\5 15 2\3 6 
Thus, (3) = ae ¢ and (23) =a = a? ; 
_1 3)\6 9 —5\12 
also, (20 254) = 6407557; and t. 8) ran id 
1 wt 
What is the m™ power of 3a2b~c?? Ans. 3%a 2 b—2me2m, 


Extraction of Roots. 


237. To extract the n” root of a monomial, extract the n” roo 
of the co-efficient, and for the new exponents, follow the rule 
given in Art. 220, viz., divide the exponent of each letter by th 
index of the root. , 

For, the exponent of each letter in the result should be such 
that when multiplied by n, the index of the root to be extracted 
the product will be the exponent with which the letter is affecte: 
in the proposed monomial; therefore, the exponents in the resul 
must be respectively equal to the quotients arising from the di 
vision of the exponents in the proposed monomials, by 2, the in 
dex of the root. ‘Thus, 


; 
oo a) aE eS 1 2 
also, a*=za?y and a°b-2 = qg>h 3, 


The rules for fractional and negative exponents have been easil1 
deduced from the rule for multiplication; but we may give a direc 
demonstration of them, by going back to the origin of quantitie: 
affected with such exponents. 


We will demonstrate implicitly, the two preceding rules. 
m 
£ hom r 
Let it be required to raise a* to the — — power. 
s 


By going back to the origin of these notations, we find that 


Je MEN AG 
(a) = Ee (Var)\evanit Sant 





1) 
| 

| 

AP. VIII.) THEORY OF EXPONENTS. 239 
‘ / 


-Remark I.—The advantage derived from the use of fractional 
xponents consists principally in this:—The operations performed 
yon expressions of this kind require no other rules than those 
itablished for the calculus of quantities affected with entire expo- 
mts. This calculus is thus reduced to simple operations upon 
actions, with which we are already familiar. 

Remark JI.—In the resolution of certain questions, we shall 
» led to consider quantities affected with incommensurable expo- 
nts. Now, it would seem that the rules just established for 
ommensurable exponents, ought to be demonstrated for the case 
which they are incommensurable. But let us observe, that the 


ilue of an incommensurable, such as a ty 11, may be de- 
rmined approximatively as near as we please, so that we can al- 
ays conceive the incommensurable to be replaced by an exact 
action, which only differs from it by a quantity less than any 
ven quantity; and we apply the rules to the symbol which 
signates the incommensurable, after substituting the fraction 
hich represents it approximatively. 


EXAMPLES. 


, 1 

/ 3 

1. Reduce Baye 519)" 
1 2 


: 1 
i 4 1 (2)2 4/3 )* 
'2.. Reduce Be) y * 


to its simplest terms. 


Ans. 44/ 3. 





T to its simplest terms. 
24/ 2 (3)? 
Lis 
Ans. ——v¥ 3. 
384 
1)3 31 )% 
,8. Reduce G) + v3 to its simplest terms. 


2/2. (3)? 
Yall 

| Ans. fob een 721). 

4. What is the product of 


5 +3 3 2 | Hs 5 
a2 + ab? + a2b3 + ab + a2b? + 53, by a? — D5. 
Ans. a? — 6. 


ol 


A 
2 


1 1 pee 
a, by a? —b %. 


te 7 Lae 1 
‘5. Divide a? —a2b 2 — a3d+5 
i 1: Ans. a? — b. 


940 ELEMENTS OF ALGEBRA. (CHAP. VIIZ 


Method of Indeterminate Co-efficients. 















238. The binomial theorem demonstrated in Art. 203, explains 
the method of developing into a series any expression of the for 
(a + 5)”, in which m is a whole and positive number. 

Algebraists have invented another method of developing alge. 
braic expressions into series, called the method by indeterminali 
co-efficients. ‘This method is more extensive in its applications 
can be applied to algebraic expressions of any nature whatever 
and indeed, the general case of the binomial theorem may be de. 
monstrated by it. 

Before considering this method, it will be necessary to explair 
what is meant by the term function. 


Let a=b+¢. 


In this equation, a, 6, and c, mutually depend on each other fo 
their values. For, 


a=b+c, b=a—c, and c=a—Qb. 


The quantity a is said to be a function of b and c, b a functioy 
of a and c, and c a function of a and b. And generally, whe: 
one quantity depends for its value on one or more quantities, it i 
said to be function of each and all the quantities on which it de 
pends. 


239. If we have an equation of the form, 
A+ Be + Ca? + Do? + Eat + &e. = 0; 


it is required to find the values of the co-efficients A, B, C, IL 
E, &c., under the following suppositions : 

ist. That no one of the co-efficients is a function of 2. 

2d. That the series shall be equal to zero, whatever be thi 
number of its terms; and 

3d. That it shall be equal to zero, whatever value may be at 
tributed to a. 

Now, since the co-efficients are independent of a, their value 
cannot be affected by any supposition made on the value of » 
hence, if they be determined for one value of a, they will b 
known for all values whatever. 


j 
ldar. VIIt.] INDETERMINATE CO-EFFICIENTS. 241 


) Let us now make 
2 =0, which gives 
! Ba + Ca? + De? + Ext + &e. = 0; 
and consequently, A=): 
Jl we divide by x, we have 
B+ Ca + Dx? + Ex? + ry =—0, 
nd by again making 2 = 0, we have 
, Ca + Da? + Ex? + &e. = 0; 
nd consequently, > SAY 










_ Dividing again by x, we have 
I i C+ De + Ex? + &e. = 0; 
nd by again making « = 0, we obtain 
| Da + Ex? + &c. = 0, 
ad consequently, Cree. O sry 
ad by continuing the process we may prove that, each co-efficient 
ust be separately equal to zero. 
It should be observed, that A may be considered the co-efficient 
ae’. 
240. The principle demonstrated above, may be enunciated un- 
xx another form. If we have an equation of the form 
@+bert+ca?+de®+ 2... = Wt Wet rt tt Wot. 
hich is satisfied for any value whatever attributed to a, the co- 
licients of the terms involving the same powers of x in the two 
embers, are respectively equal. For, by transposing all the terms 
to the first member, the equation will take the form 
( A+ Bu + Ca? + Dx? + Ext + &e. = 0; 
Bence, a—a—0, b—b'=0, c—c’=0....; 
id consequently, 
foam Cee c,d Bad ee 

Every equation in which the terms aie atranged with reference 
a certain letter, and which is satisfied for any value which 
ay be attributed to that letter, is called an identical equation, in 
‘der to distinguish it from a common equation, that is, an equa- 
m which can only be satisfied by particular values of the un- 


20wn quantity. 
16 


242 ELEMENTS OF ALGEBRA. [CHAP. VIII 


241. Let us apply the above principles in developing into » 


series the expression 
a 


a+ Ve 

It 1s plain, that any expression equal to the above, must cor 
tain a, and the quantities a, a’, b’. Let us then assume 

Taye At Bet Cx? + Dai + Eat + &e. . .. (1), 
in which the co-efficients A, B, C, D, &c., are functions of a, a 
Y, and independent of w. These are called, indeterminate co-ef 
cients. It is required to find their values in terms of a, a’, b’, ¢ 
which they depend. 

For this purpose, multiply both members of equation (1) | 
a’ + b’a. Arranging the result with reference to the powers | 
x, and transposing a, we have 

nee Ad’ + Ba’ |x + Ca’ |x? + Da’ | a? + Ea’ 

~ (-—a +A} +BY! +CH + Db' 
and since this equation is satisfied for any value of 2, we ha 
(Art. 239), 


at + vi 2 














Aad’ —a=0, whence, As; 
a 








also, Ba’ + Ab’ =0, whence, 
pe Ae 
a’ a a af? 
also, Ca’ + Bb’ = 0, whence, 
Bb’ ab’ bf ab’? 
on — Bia. (aa 
also Da’ + Cl’=0, whence, 
CY ab” b al/3 
D= — — = 3 \* — nn &c. 


It is plain that .ue terms will be alternately plus and min 
and that the co-efficient of any term is formed by multiplying ¢ 


b’ 
of the preceding term by — ee therefore, we have 


a a ab’ ab’2 ab’3 ab/4 
ol Poaieetadipeashes NPE pe 


—_—_—_ = x at — &e. 
a + b'x aera a’3 a’4 a’ 








" 


AP. VIII.) INDETERMINATE CO-EFFICIENTS. 243 


142. The method of indeterminate co-efficients requires that we 
uld know the form of the development. The terms of the de- 
opment are generally arranged according to the ascending powers 
“, commencing with the power x°; sometimes, however, this 
n is not applicable, in which case, the calculus detects the 
or in the supposition. 


2 1 
“or example, develop the expression rae 


uet us suppose that 


3a — x2 i" 


mee, by reducing to entire terms, and arranging with reference 
e, 











PO = — 14 3Ar+ 3Ble2+3Cl]2243Diat+... ee 
—A —B —C 
mee (Art. 239), 
pes, S60 0,1) BB 014 & Op yoy, 
low, the first equation, —1—0, is absurd, and indicates that 
above form will not develop the expression ee But if 
ie — 
put the expression under the form it * er and make 
= a 
gg At Bet Ca? + De? + ..., 
shall have, after the reductions are made, 
9 — $344 3B lr+4+3Cla2+3Dla+ ..., 
an } —1—A —B —C 











ch gives the equations 
Pree eRe ao. 50 2 Bg 


14 1 1 1 
= — B= — = ae EE she 6 
Me. si 8? 9” 27’ 81 
herefore, 
) aoe =~(- 1 oer ee ) 
We Sy gt. Te gg ge tgp t -.) 
, 


] 1 l i; 
eae LS —] — 79 — —— Co arte 
mae ia om | 97% aise ictal : 


is, the development contains a term with a negative expo- 





244 ELEMENTS OF ALGEBRA. (CHAP. VII 


Recurring Series. 


243. The development of algebraic fractions by the method o 
indeterminate co-efficients, gives rise to certain series, called 7 
curring series. 

A recurring series is the development of a rational fraction u 
volving x, made according to a fixed law, and containing the ascen 
mg powers of X in its different terms. 


It has been shown in Art. 241, that the expression 











a a ab/ ab’ ab’? 
Te Lida” | 
in which each term is formed by multiplying that which preced 
b’ 
it by eras 
a 


This property of determining one term of the development frc 
those which precede, is not peculiar to the proposed fraction; 
belongs to all rational algebraic fractions, and may be thus ¢ 
pressed ; viz., Every rational fraction involving x, may be develoy 
into a series of terms, each of which is equal to the algebraic st 
of the products which arise from multiplying certain terms of a p 
ticular expression, by certain of the preceding terms of the series 

The particular expression, from which any term of the ser 
may be found, when the preceding terms are known, is called | 
scale of the series; and that from which the co-efficient may 
formed, the scale of the co-eflicients. 


: : b’ Be: 
In the preceding series, the scale is — —a, and the series 
a 3 | 
b’ a 
called a recurring series of the first order; and yar is the sc 


of the co-efficients. 
a+ be 


_1- et It Hh SOTIENS 
a + Wa + cx? 


Let it be required to develop 


Assume 
a+ bx 
d+ Va + cx 
reducing to entire terms, and transposing, we have 
Ad + Ba’ \x2+ Ca’ | a? + Da’ |a?+ Ea’ |a*+ . 
0o—~ 2 —a+Av| +BY); + Ci’) + DY 
— b\ + Ac) + Be) = ee 


= A+ Br+ Cx?+ Dx + Ext+ .. 





s 
! 
! 







‘AP. VIII.) INDETERMINATE CO-EFFICIENTS. 245 


ich gives the equations 


H Ad —a=0, or A=-—, 


a 
/ 

Ba’ + Al’ — 6 =0, whence, Be aie edt dees 

a 

; vy re 
Ca’ + Bb’+ Ac’ = 0, whence, er hak ity ists 

a 

| bY ce 
— Da’ + Cb’+ Be =0, whence, LRA tC eS oad, 

a 

: \ b’ Cc 
Ed + Dl’ + Ce’ =0, whence, E=——D——C. 

a 


ym which we see, that the first two co-efficients are not ob- 
aed by any law; but commencing at the third, each co-efficient 
‘formed by multiplying the two which precede it, respectively, 









b/ oe. aga 
‘—— and —-—, viz., that which immediately precedes the 
(me @ a 
by 
uired co-efficient by — —, that which precedes it two terms 
a 


Se ; 
/—-—, and taking the algebraic sum of the products. Hence, 
ha 


b’ c’ 
(— n 2) 


the scale of the co-efficients. 
From this law of the formation of the co-efficients, it follows 
t the third term of the series, Cx?, is equal to 


b’ 
— — Ba? — — Ax’; 
/ / 
a a 
b’ c’ 
r ——x.Br——2*.A. 
a’ a’ 


n like manner, we have for Dz’, 


b’ Cc’ 
— — Cz3 — — Bz'; 
/ / 
a a 
b’ Cc 
r ie ey? — 2. Bo. 
a a 


dence, each term of the required series, commencing at the 
d, is obtained by multiplying the two terms which precede 


246 ELEMENTS OF ALGEBRA. [CHAP. V) 


respectively by 


and taking the sum of the products: hence, this last express 
is the scale of the series. 

Recurring series are divided into orders, and the order is e 
mated by the number of terms in the scale which involve 2. 


: a 
Thus, the expression ———- gives a recurring series of 
wa a + Wx 
& Eh te b/ 
first order, the scale of which is — — @. 
a 
a+ bx 





The expression gives a recurring series of 


a’ + b’x + cx? 


second order, of which the scale is 
/ 


( b’ c 2) 
a— By Sits $ 
a a 


The series obtained in the preceding Art. is of the sec 


order. 
In general, an expression of the form 








at be + cx? ou. . hae 
Gt Wetec +... eae 
gives a recurring series of the n' order, the scale of which - 
b’ ie k/ 
—-—S%4 -—- Tae. . =o eT 
a a a 





Remarx.—lIt is here supposed that the degree of a in the 
merator is less than it is in the denominator. If it was no‘ 
would first be necessary to perform the division, arranging 
quantities with reference to 2, which would give an entire < 
tient, plus a fraction similar to the above. 

1 — a — 3x? + 4a3 4+ at 
 2— 5a + 327 — a 


at + 403 — 3a? -—a4+ 1 — + 32? —5r4+2 


Thus, in the expression , gives 


+ Ta? — 8x? + 2 |) 
18a? — 340 + 15. 
Performing the division, we find the quotient to be — # - 
plus the fraction 
13a? — 34a + 15 15 — 34a + 132? 


Tia | Be — Sa +2 2— 52 +322?— 2? 





‘HAP. VIII.] BINOMIAL THEOREM. 


247 


Demonstration of the Binomial Theorem for any Exponent. 


\ 244. It has been shown (Art. 61), that any expression of the 
,orm | 











a” — y™, is exactly divisible by x — y, 


when m is a positive whole number. ‘That is, 
t am a #5 y™ 


| ce . y = yr + amy + rae SE Oe _. nye? iC y™)~ 





The number of terms in the second member is equal to m; and 


f we suppose «= y, each term will become equal to x’: 
lence, 





CY ieee 1 


We propose to prove that the quotient will have the same form 


when m is negative, and also, when m is a positive or negative 
fraction. 


First, when m is a whole number, and negative. 
| Let x be a positive whole number, and numerically equal to m 


Then, 





m—=—n. 


By observing that 


=H aay” x (ar Pes, y”) — grr ag 


ive have 
oy an nm an 
C y _ — lee Sa x ben ox) = —2nyn pn = nz}, 
L—y oy 
: 
ter making y= 2; and by restoring m, 
| a om — xm 
|| ° ——_—— = mx"—1, m being a negative number. 
f Z2—w 


Second, let m be a positive fraction, or m P 


—_—_— —— 
— e 





p 

Let a%==v, whence, wY¥=v?, and x=v!; 
i 2 

and y?=u, whence, y{=u?, and y= w!. 
pay aa eR 

an?’ — yf vP — uP v—U 
Hence, 8) saltgpas |} cetnnilaal 
- irae vi — uf vi — uf 


v—wu 


248 ELEMENTS OF ALGEBRA. [CHAP. VIII 


If we suppose = y, we have vy =u; and since p and g art 
positive whole numbers, we have 
= 2 
a’ — rf ppl =—} 
x— x qui} q q 





after substituting for v its value a. 
If we restore m, we have 





a™ ee am 
——— = mx™—!,. m being a positive fraction. 
L—e2 
Third, let m be a negative fraction, or m= — f. 
q 
es we 
Let 2?==v, whence, « 7=v-?, and # = 0%; 
at 
and y2=u, whence, y%=wu-?, and y= uf. 
MURS 2 bie Ge? 
of v ?P —uP v—U 
Hence, 2 ede i — 
v—y vf — us vi — uf 
v—U 


But since —p is a negative integer, and g a positive integer 
we have from what has preceded, after making x = y, whict 
gives u =v, : 




















a 2? 
ei—ea?7 — pu P-l 
£ so ees Je sat, 
vi & qe q 
- | 
and substituting for v its value #?% we have | 
Bay ign, 
xai—a 7 ee 
=—pv Pi= ——xwe7 ft . 
x— 2 R q 


and restoring m, we have 


16 fame : A : 
——— = mx™"|, m being a negative fraction. : 
x—er. 





245. We are now prepared to find a general formula for th 
development of the binomial (a + 6)", in which the exponent m 1é 
positive or negative, and either integral or fractional. 

In order to simplify the process, let us place the binomial 








f 
\)HAP. VIII.] BINOMIAL THEOREM. 249 
inder the form 


(a+ "= fa(1+—) J" =o (1 +2)” (an. 220). 


If we find the development of (1 + aay and then multiply 


by a™, the product will be the development of (a+ 5)”. 
| In order further to simplify the expression, let us make 

serene a he 

a 
aen, the binomial to be developed will be of the form 

(1 + 2)”. 
| As this development must be expressed in terms of x, and known 
‘uantities dependent for their values on 1 and m, we may assume 
| . x 


(1 + a)” =A-+ Bu+ Cx? + Dx? + Ext + &e. . . . (1), 
1 which the co-efficients A, B, C, &c., are independent of a, 


nd functions of 1 and m. 


_ Now, since this equation is true for any value of a, if we make 
'= 0, we have 





(Lye se Abe). 
_ Substituting this value in equation (1), we have 
(1+ 7)™=1-+4 Ba+ Ca? + Do? + Ext + &e.... (2). 
Since the form of the above development will not be changed 
y placing y for x, we may write 
| (1+ y)™"=1+4 By+ Cy aE Dy? sp Byes tcc: Seay, 
Subtracting equation (3) from (2), and dividing both members 
y «—y, we have 
m __ m — 72 = 
tee)" (1 +9)" =p) Y) oe =) 4p =) 
ale ey 
Make 1+ 4=2, and aa Brey e—y=v—u. 


Substituting these values in the first member of equation (4), 
nd we have 


+e... (4). 


a = peo) C294 ERY 4 peas. ACG. aan (O)e 


v—u 2—y x — 
If now, we make 
3 2 =y, whence, v= 4, 


250 ELEMENTS OF ALGEBRA. [CHAP. VIII 


we have, from Art. 244, 


om — ym ym — y™ 
Paes mre = ‘i se = mu™—1 =m (1 oa 0) erga 





while the quotients in the second member become, respectively, 














im 2 ee 
‘a Le J = 2227! = Qa5 
x—y x—y 
3°58 aS: 
= 7 x= Soto] = 3a? : POT = 4gt4 = 493 1 dic. 
xy e—y 
Substituting these values in equation (5), and we have 
Ls te Rye at B+ 2Ca + 3Da? + 4Ex? + &e..... (6). 


Multiplying both members of this equation by 1 +2, and ai 
ranging the second member with reference to «, and we have 
m(1+0)"=B+2C a+ 3D|e2+ 4E |a*+ &e. 
+ Bl +2C + 3D 











If we now multiply equation (2) by m, we have 
m(1 + «#)™=m-+ mBax + mCx? + mDx? + mExt + &e. 
If we place the second member of the last two equations equ 
to each other, we shall obtain an identical equation. Then, placit 


the co-efficients of the like powers of # equal to each oth 
(Art. 240), we have 


B=m, whence, == 

2C+ B=mB, whence, La@e)) ee 
2 Is a 

3D+2C=mC, whence, Dot) es? 

. 3 1). 2 eee 

4E+3D=mD, . whence, p_D(m—3) _m(m—1) (m—2) (m—3 

4 1, 5, Soe remem fa 
&c., &c. 


Substituting these values of A, B, C, D, &c., in equation (+ 
we obtain 
m(m—1) 


(l+2)"=l+mt+7—s eh 


“a 2 eos 


m(m — 1) (m—2) (m 


Wik eae em 


nett 
ay ae at + &e. 


Ba) tes 


’ CHAP. VIII.] BINOMIAL THEOREM. 25) 


If we now replace z by its value ih we have 
a 


m(m — 1) 6% | m(m—1)(m—2) 0 


1) 42 ro Ta Ne! ad aT oP ae ae 


Finally, multiplying by a”, we obtain 


(a + 6)" = an + ma™™b + a ath 


ENGEL ED 2) ag rare. 


ES aa 3 


f 
a development which is of the same form as the one obtained in 


: 


_ Art. 203, under the supposition of m being a positive and whole 


_ number. 
Applications of the Binomial Theorem. 
If in the formula ; (# + a)™ = 
a’ m—l1 a m—l m—2 @& 
| @ eee mS tm. a a ) 


: Sete ZL 
we make m= —, it becomes (x + a)” or A yg Se 
n 


" 

















1 1 1 1 9 

i Fe jd Me Seg?!) 1m an. 

A ee ee 
Z ( ie an HA cation 2 3 js Aas 
or, reducing, "/ «x + a= 
| eee 7 ae n—1 2n—1 a 

Le Rey ) 
4 i+— x EET ein 2n 3n x3 


_ The fifth term can be found by multiplying the fourth by 
3n — 1 





and by < then changing the sign of the result, and 


2 


so on. 

Remarx.—lIf in this formula, we make n= 2, n= 3, n= 4, 
é&c., the development will become the approximate square root, 
cube root, fourth root, &c., of the binomial («+ @); and by as- 


signing values at pleasure to x and a as well as to m, we can 


coal 


find any root whatever of any binomial. If n is negative, or frac- 
tional, there will be no limit to the number of terms to which the 


series may be carried. Such a series is called an infinite series. 


252 ELEMENTS OF ALGEBRA (CHAP. VIII. 


The binomial formula also serves to develop algebraic expres- 
sions into series. 


: 1 
Take, for example, the expression i= owe have 
1 
ot ee) eee 
aera ge 
In the binomial formula, make m= —1, r=1, and a=— a2; 
it becomes 


(1 2) = 111 (+ 2) fel ge 





or, performing the operations, and observing that each term is 
composed of an even number of factors affected with the sign — 


(Q— ays Slt et Pt eet tert fees = 
—z 


The same result will be obtained by applying the rules for di- 
vision (Art. 55). 
1{|1l—2 
Ist remainder - +2/1+2+22+ gh ee eee 
Gd ee ee 28 | 








30-9 - «- 1s He 
4th -+- .- = +t 
Ee hay tae 
2 
; ‘ hol. aan Ue peat 
Again, take the expression (2 or 2(1 —2z) 
We have 2(1—2z)3= 
ce ad —3—1 —3=2 

—3.(—z)— —z)?—3. .— >= .(—2z}?— - J; 

2[1—3 .(—z)—3. 5 .(—z)?—3 5 5 (—z) ] 


or = - 2 (1 —z)3=2(1 + 32 + 62? + 10294 1524+. . .) 
To develop the expression ‘/2z — z*, which reduces to 


iL 
4/22 (1 —)?, we first find 








z\ il 4) 1 bog aN 
—_ &s — |] _— ped Peneaeete = © . Te te ea - ORS 
(1 =) 3 4 Tr 3 2 P.) me 
1 1 5 
DE es, RANE iy ae Pt ae 
‘ARR, 648 ~ 


. CHAP. VIII.) BINOMIAL THEOREM. 253 


a 1 5 
- hence 22 — 22 = V 22 n(loas sis at — — 2 —, &e,) 


EXAMPLES. 


1. To find the value of cas (a + b)-? in an infinite 
series. 


CF i Firs) ; 
2. To find the value of Bs a8” in an infinite series. 
' r+¢e 


x? x3 at 


Ans. r—x2+———+ —, &e. 
r roe howe 
; atx? Pate 5 
3. Required the square root of ———— in an infinite series ~\ 
a — x 
6 
. Ans. L+5 oe Se a8 &c. 
v 7 2 
4. Required the cube root of ie in an infinite series. 
1 2x? 5a* 40x® 
7 9 igga i aa a &e.) 
‘1 as s 3a? 9at Blas’ “~ 


_ Remarx.—When the terms of a series go on decreasing in 
| value, the series is called a decreasing series ; and when they go 
| On Increasing in value, it is called an increasing series. 

_ A converging series is one in which the greater the number of 
terms taken, the nearer will their sum approximate to the true 
value of the entire series. When the terms of a decreasing and 
converging series are alternately positive and negative, we can, by 
taking a giyen number, determine the degree of approximation. 

' For, let a—b+c—d+e—ft. , &c., be a decreasing 
| Series, b, c, d,... being positive fadneheies, and let « denote the 
| true value of this series. Then, if » denote any number of terms, 
the value of « will be found between the sum of the n and n+1 





+ For, take any two consecutive sums, 
@a—t+e—d+e_f and a—b+c—d+e—f+g. 
In the first, the terms which follow —f, are 


fp + g—A, +k—-I+...; 


254 ELEMENTS OF ALGEBRA. (CHAP. VIII. 


but since the series is decreasing, the differences of the consecu- 
tive terms g—h, k—Il,... are positive numbers ; therefore, in 
order to obtain the complete value of x, a positive number must 
be added to the sum a—b+c—d+e-—f. Hence, we have 


a—bte—d+e—f<cu. 

In the second series, the terms which follow +g, are —h+4&, 
—ltm.... Now, the differences —h+hk, —l+m..., 
of the consecutive terms, are negative; therefore, in order to ob- 
tain the sum of the series, a negative quantity must be added to 


a—b+c—d+e—f+g, 


or, in other words, it is necessary to diminish it. Consequently, 


a—btc—d+e—f+g>u. 


Therefore, 2 is comprehended between the sum of the n and n+1 
terms. 

But the difference between these two sums is equal to g; and 
since x is comprised between them, their difference g must be 
greater than the difference between a and either of them; hence, 
the error committed by taking n terms, a —b+¢e¢— d+e—f, of 
the series, for the value of x, is numerically less than the follow- 
ing term. 


Summation’ of Series. 


246. An interesting, and at the same time useful application of 
the principles involved in the summation of series, is found in 
determining the number of balls or shells contained in a given pile. 

Let ABC be a triangular pile of balls, 
having eight balls on each of the three 
equal lines, AB, BD, and AD, and also, 
eight balls in height along the line CB. 

Now, the proposed pile consists of 8 
horizontal courses, and the number of 
shot in each course, is the sum of an 
arithmetical series of which the first term 
is 1, the last term the number of courses 
from C, and the number of terms, also 
the number of courses from C. There- 





“HAP. VIII.) SUMMATION OF SERIES. 255 


ore we have 





Ist course is equal to (1+1)x 4= 1; 
2d (73 66 (1+ 2) x1 — 3 
eo gent cet! (1+3)xlk= 6; 
4th « ‘ (1+4)x2 =10; 
Sth |“ ‘ (145) x 2h= 15; 
" Bibs es (1 +6) x3 = 21; 
Th “ (1 +7) x 31 = 28; 
8th « “ (1+8)x4 =36. 


| 
Hence, the number of shot in the pile will be equal to the sum 
pe the series 

/ 


1, 3, 6, 10, 15, 21, 28, 36; 


i which any term is found by adding 1 to the number of the term 
‘md multiplying the sum by half the number of terms. 
. Thus, if we suppose the horizontal layers to be continued down, 
(md denote the number of any layer from the top by n, we shall 
jave 
m(n-+ 1). 

D at, 
ud the sum of this series will express the number of balls in a 
riangular pile, of which n denotes the number in either of the 
»ottom Tows. 
If the general term of any increasing series of numbers involves 
2 to the m” degree, the sum of the series will not involve n to 
1 higher degree than (m+ 1). For, the sum of such series can- 
10t exceed n times the general term, and hence, cannot involve 
ato a higher degree than m+ 1. Let us therefore assume 


n(n +1) 
2 





1s. 10,.15, 21, 


ee 3 + Garo 15-8." — A+ Bn + Cn? + Dni, 


n which the co-efficients A, B, C, and D, are not functions of n. 
In order that these co-efficients may be determined, we must find 
four independent equations involving them. If we make 

a=1, we have A+B+C+D=1 e=) LwGhh 
i2=2, gives A+ 2B4 40+ 8D=1+3 4a Gap 
ime 3, © A+3B+-9C0+27D=1+4+3+4+6 2.1093), 
m4." A+Y4B+16C+ 64D=1+3+6+4+10=20 (4). 





256 ELEMENTS OF ALGEBRA. (CHAP. VIl) 


Now, by a series of subtractions we have 
Equation (2)—(1), gives B+ 3C+ 7D= 3... (5), 


« (3)—(2), © \-Bre SO 19D = (Gua), 
«  (4)—(3), “ B+7C+37D=10... (7), 
PS (CO a aes BC + 12D = 9B ete), 
Se fc) e629) 2C+18D= 4. . 2(9), 
1 
(9)—(8), 6D = 1; hence, D == 
also, 2C+18D = 4, gives => 
B+7C+37D=10, « B=— 
A+B+C+D=1, a! A = .03 
Hence, 
1 
143¢ot10¢ 194+... 2 FD pnt det + 3 


== (2+ 3n + 2”) 


+ 1) (2+ 2) 


were as 


. 
~ — 


He n(n 
1 


Let us suppose that we 
have a pile of balls whose 
base is a square, two sides 
of which, EF, FH, are seen 
in the figure, and that it 
terminated by a single ball 
at G. 

Now, the number of balls 
in the upper course will be 
expressed by 1?, in the sec- 
ond course by 27, in the 
third course by 3%, &c. Hence, the series 

12 4+ 92 + 93 6/42 - 52 
will express the number of balls in a square pile, of which th 
number of courses, and consequently the number of balls in on 
of the lower rows is, n. 
To find the sum of this series, assume 


14+44+94+164+ ...n23=A+ Bn+ Cn? + Dri, 





* 






HAP. VIiL.] _ SUMMATION OF SERIES. 257 


|om which we find 


A+2B+ 4C+ 8D=1+44 Serums 
A+3B+ 9C0+27D=1+4+4+9 roel 


A+4B+4+16C+ 64D=1+44+9+416=30; 

ad from these four equations, we find, by continued subtractions, 
ae = 5, D— +, and A=0; hence, 

P 1444+9416425..... n® = 1n + 1n? + 1n3 

= = (2n? + Bn + 1) 

| __n(n +1) (2n +1) 

| Pe AC MINN Hie 1) 


| 





‘ 
Let us now suppose that we have a rectangular or oblong pile 
f shot, as represented in the figure below. 








Suppose we take off from the oblong pile the square pile FFD. 
Ve then see that the oblong pile may be formed by adding to 
je square pile a series of triangular strata, each containing as many 
alls as are contained in one of the faces of the square pile; and 
xe number of the triangular strata will be one less than the num- 
er of balls in the top row. Therefore, if n denote the number of 
orizontal courses, the number of balls in one triangular strata 


n(n +1) 
2 






rill be expressed by - and if m+1 denotes the whole 


jumber of balls in the top row, the number of triangular strata 
vill be denoted by m; and the number of balls in all these strata 


, et) sem, 


17 * 


258 ; *  BLEMENTS OF ALGEBRA. (CHAP. VII 


But since the number of balls in a square pile, whose side con 
tains n balls is 
n(n + 1) (2n+ 1) 
LO.) ae nls 
the number of balls in an oblong pile, whose top row contain 
m-+1 balls, and depth n balls, will be expressed by 
n(n + 1) Coane ye n(n + 1) 


—_—* x 


ori fons 3 2 
_ n(n+1) A +2n-+ 3m) 
amet 


Sa 
If we denote the general sum by S, we shall have the follov 
ing formulas for the number of shot in each pile. 








vatn 1) Oe) ae n(n + 1) 
Triangular, alee ee A (n+1+1 
CO, CS oct) 27S aoe | Yb n(n + 1) 
Square, oS ae rT eee ha ea 
Rectangular, 
1) (2n+-1+3 ] 1 : 
sat PE ge aR ) Cnt) + (m-tin)-+ (m+ ip 


Now, since a ae 


is the number of balls in the triangul: 
face of each pile, and the other factor, the number of balls in th 
longest line of the base plus the number in the side of the bas 
opposite, plus the parallel top row, we have the following 


RULE. ; 
Add to the number of balls in the longest line of the base, t: 


number in the parallel side opposite, and also the number in the ti 
parallel row; then multiply this sum by one third the number in 
angular face. 

EXAMPLES. 


1. How many balls in a triangular pile of 15 courses? 
Ans. 680. 


2. How many balls in a square pile of 14 courses? and ho 


many will remain after 5 courses are removed ? 
» . Ans. 1015 and 960.’ 
f ’ e 


i 

' 

HAP. VIII.] SUMMATION OF INFINITE SERIES. 259 

' 

8. In an oblong pile the length and breadth at bottom are respect- 
‘ely 60 and 30: how many balls does it contain? Ans. 23405. 
4. In an incomplete rectangular pile, the length and breadth at 
ttom are respectively 46 and 20, and the length and breadth at 
p 35 and 9: how many balls does it contain? Ans. 7190. 


4 Summation of infinite Series. 


247. An infinite series is a succession of terms unlimited in 
mber, and derived from each other according to some fixed and 
own law. 

‘The summation of a series consists in finding an expression of 
finite value, equivalent to the sum of all its terms. 

Different series are governed by different laws, and the methods 
‘finding the sum of the terms which are applicable to one class. 
Il not apply universally. A great variety of useful series may 
- summed by the following formula : 











| q q PY 
Assume =—— Bann 
| n n+p n(n-+p) 
then, es Jai (2 — —f_), 
; n(nt+p) p\n n+p 
Tf now, by attributing known values to p and g, and different 
: : i qg 
ues In succession to n, the expression —— shall repre- 
[ 3 n(n + p) 4 


it a given series; then, the sum of this series will be equal 
2 . 
,— multiplied by the difference between the two new series 





which % and —2— are the general terms. Hence, if the dife 
n n-+p 


ence of the sums of these series be known, and the value of 
q 





‘be known, we can find the value of the series ————. by 
Re wT ip) 

formula S = ol (s‘—s’) even if we do not know the value 

P 

the new series and —2—. 

| ne n+p 

| EXAMPLES. 

'. Required the sum of the series 

ie 

ie .! ++ > aera ue + &c., to infinity. 





1 Me wes” 1.3.4 
i & 


260 ELEMENTS OF ALGEBRA. (CHAP. VIII. 


_ We see that if we make g = 1, and p= 1, and n= 1, 2, 3. 
4, &c., in succession, that the first member of the formula, 


q 
n(n + p)’ 
will, in succession, represent each term of the series; while un 
der the same supposition, the second member will become, for : 


terms of the series, 








hci aps 1 
2 3 An a ae n 1 1 n 
1 1 a 1 = Te aa tT 
\csusamepane’y ott 


If now, we suppose n= , the value of the sum of the serie 
will become equal to 1. F 
2. Required the ‘sum of n terms of the series 


1 1 1 1 1 uber 
ra kaieR ch sa} 7a9 T oo pay: 


('o adapt the formula to this series, we make g=1, p= 
and n= 1, 3, 5, 7, &c.; we then have, for the sum of n term 


pep ae : 
Ee Ba i ieee , 1 
| pe sp 1 1: eae be 
| Greate Se nh ae 
2 
Bei Me acct and- he of this sum = eee 
Qn +1 p Qn+ 1 


If now, we suppose n=, the value of the series becom 
equal to one half. 
3. Required the sum of n terms of the series 





1 1 1 1 

Joy 1 ee 

iia hare tu 77 Pr eee 

Here p= 3, g=1, n=1, 2, 3, 4, é&c.: hence, 
Po a i 

1 a 7.3. | ee n 

ey Rees fy 1 1 1 1 
<(¢+5+4 "a "aed Dee 








f 
| 


CHAP. Vit.) . SUMMATION OF INFINITE SERIES. 261 


4. Find the sum of n terms of the series 








Ee 2s. Oa ye nL oe 
3.5 «5.7 7.9 9.11 Trg t &e» 10 infinity 
were aes get] 

ieee Goet re | 
| a en Ta = n-+1 
| Seer i: 5 Intel 7.2n 8 


vhich becomes, 
2 alee + 1 


| oe rt et eee + 1) 
If the number of terms used is even, the upper sign will apply, 
ie quantity within the parenthesis will become + 1, and the sum 


{the n terms before dividing by p, is 
1 n+ 1 


———~ = —, when n=o. 

toa en 3 6’ 

| If n is odd, the lower sign is used, and the quantity within the 
arenthesis reduces to zero, and we have 


— — ———_- = —, when n= op. 


‘Then, since p = 2, the sum of the series when n= o, is 2 
5. Required the sum of the series 
t 4 4 4 4 4 


5 5.9 9.13 + 13017 + T7737 
& Ans J 





+ &c., to infinity. 





262 ELEMENTS OF ALGEBRA. [GHAP. L 


CHAPTER IX. 


CONTINUED FRACTIONS, EXPONENTIAL QUANTITIES, LOGARITHM: 
AND FORMULAS FOR INTEREST. 


248. Every expression of the form 


ite ee bee 

c T ae 

in which a, b, c, d, &c., are positive whole numbers, is calle 
continued fraction. 

Hence, a continued fraction has for its numerator the unit 1, 
for its denominator a whole number, plus a fraction which has 1 
its numerator and for its denominator a whole number plus a f 
tion, and so on. 

249. The resolution of an equation of the form 

a* = §, 
gives rise to continued fractions. Suppose for example, a= 
b = 32. We then have 


g7 — 32, 

in which 2 >1, and 2<2. Make 
1 
e=-1+-—, 

y 


in which y > 1, and the proposed equation becomes, after cl 
ging the members, 


32—8 %™=8 x 8¥, whence, 
1 
8¥—4 and consequently, 8 = 4’. . 


CHAP. IXx.] CONTINUED FRACTIONS. 


263 
It is plain, that the value of y lies between 1 and 2. Suppose 
1 
tie 1 -+- -o 
1+ & 
and we have, Biome) * == 4 & 44; 
1 
hence, 4*—2, and 4=2*, or z= 2 
1 1 3 
B t = 1 —_ — —> — ——:: 
ul, y 7 on EO 5) 3” 
; 1 ] 2 5 
and neat | —-—]1 —=1 ae oe, ee 


and this value will satisfy the proposed equation. For 


gr = 88 = 9/85 = 3/(29)5 = 3/(24)8 = 28 = 32. 
250. If we apply a similar process to the equation 
; 10% = 200, 
we shall find 
1 1 1 l 
e=2+—; ai 5 2=3+—; “u™—3+—. 
‘ y Zz U i 


Since 200 is not an exact power, x cannot be expressed either 
by a whole number or a fraction: hence, the value of x will be 
incommensurable, and the continued fraction will not terminate, but 
will be of the form 


i 


1 
as cas) 
an u+ &c. 


_ 251. Common fractions’ may also be placed under the form of 
continued fractions. 











i 65 Nat : 
Let us take, for example, the fraction 140’ and divide both its 


rms by the numerator 65, the value of the fraction will not be 


shanged, and we shall have 1 
« NO’ Neinct, ie 
Rs 65 149 
149°” «65 
ae One 65 1 
| effecting the division, a ao 


264 ELEMENTS OF ALGEBRA. (CHAP. IX. 


LO): x 
Now, if we neglect the fractional part a5 of the denomiuator, 


we shal] obtain _ for the approximate value of the given frac- 


tion. But this value would be too large, since the denominator 


used was too small. 


iY) 
if, on the contrary, instead of neglecting the part gg? ve Wete 


1 
to replace it by 1, the approximate value would be 3” which 


would be too small, since the denominator 3 «is too large. Hence, 


Macias : 1 ] 
therefore the value of the fraction is comprised between ba and z 


If we wish a nearer approximation, it is only necessary to op- 


; 19 } ; ; 
erate on the fraction az a8 we did on the given fraction 





149” 
and we obtain ’ 
19: )jobped 

65 © 8 
3+ —, 
19 
65 1 
hence, SOC Bs eal: 
2+ 
ue 
19° 


8 ° ‘ 
If now, we neglect the part Ty the denominator 3 will be less 


be ad 
than the true denominator, and 4 will be larger than the num- 


ber which ought to be added to 2; hence, 1 divided by 2 | 


will be less than the true value of the fraction; that is, if we sto} 
at the first reduction and omit the fractional numbers, the result 
will be too great; if at the second, it will be too small, &c. Hence, 
generally, if we stop at an odd reduction, and neglect the fractional 
part, the result will be too great; but if we stop at an even reduc- 
tion, and neglect the fractional part, the result will be too small. 


Ws 
t 
| 
HAP. IX.) CONTINUED FRACTIONS 265 


Making two more reductions in the last example, we have 
35 1 




















: [ae 
| 2-+ Pathe 2. Ist reduction, too great; 
3+ , : 2d ge too small ; 
ach 3d a too great; 
2+ 4th too small; 
1 + ct 5th e too great. 
 252..The separate fractions 
25 1 1 
: peatiker. 7 
Sipe ats, tee se 
b+ —, 
c 


re called approximating fractions, because each affords, in suc- 
ession, a nearer value of the given expression. 


Pag) 
_ The fractions —, —, —, &c., are called integral fractions. 
) ae & Cc 


When the expression can be exactly expressed by a vulgar frac- 


on, aS in the numerical examples already given, the integral 
ee | 


Se &c., will terminate, and we shall obtain an 
c 


xpression for the exact value of the given fraction by taking 
rem all. 


‘ 1 
‘actions —, 
a 


We will now explain the manner in which any approximating 
‘action may be found from those which precede it. 











1 1 ' : 
; 1. a hie tess le oe Ist app. fraction. 
BE’ 1 b : 
2. j a aT 2d app. fraction 
a+ — 
b ‘ 
i? 1 be+ 1 hs: 
; So a eee: | 
3 ; Gipije es app. fraction 
| 1 
| ee 
c 


By examining the third approximating fraction, we see, that its 
umerator is formed by multiplying the numerator of the prece- 
ing fraction by the denominator of the third integral fraction, and 


\ 


266 ELEMENTS OF ALGEBRA. (CHAP. Ix 


adding to the product the numerator of the first approximating 
fraction: and that the denominator is formed by multiplying the 
denominator of the last fraction by the denominator of the third 
integral fraction, and adding to the product the denominator of 
the first approximating fraction. 

We should infer, from analogy, that this law of formation is 


Q 


general. But to prove it rigorously, let PO RB be any three 


approximating fractions for which the law is already established. 
Since c is the denominator of the last integral fraction, we have 
from what has already been proved, 


R ve Qc + P 
Ro. We ee 


: 1 
Let us now add a new integral fraction a to those already 


S BONY 
reduced, and suppose -— to express the next approximating frac 


SS’ 
; Pee S mT 
tion. It is plain that 7% will become v by simply substitutin: 


for c, c+ =: hence, 


8 2+ G)+P act PyatQ Rd+Q 
(ri) 








Suen 





~(Qe+P)d+ QV RAt+ QV. 


Hence, we see that the fourth approximating fraction is deduce 
from the two immediately preceding it, in the same way that th 
third was reduced from the second and first; and as any fractio 
may be deduced from the two immediately preceding in a simila 
manner, we conclude that, the numerator of the n™ approwimatin, 
fraction is formed by multiplying the numerator of the precedin 
fraction by the denominator of the n™ integral fraction, and addin 
to, the product the numerator of the n — 2 fraction; and the denon 
inator-is formed according to the same law, from the two precedwy 
denominators. 


253. If we take the difference between any two of the cor 
secutive approximating fractions, we shall find, after reducing thet 
to a common denominator, that the difference of their numeratot 


ee 


‘, i 
% ‘ *. sy 


‘OH AP. IX.] CONTINUED FRACTIONS. 267 


will be equal to +1; and the denominator of this difference will 
be the product of the denominators of the fractions. 


; { ; 1 b 
Taking, for example, the consecutive fractions pre and peas 
we have 
1 eee hu, 80 cpa 20 Fy od! tae I 
@ ab+1 a(ab+1) ~~ a(ab+1) 
b be + 1 aig | 
and 


SMR ea Dato, DL ah-nhietal 
To prove this property in a general manner, let 
PUP rantly 
PR Geer 
be three consecutive approximating fractions. Then 


Rai a a ays ea th 








iE PN RIE BOM 
yi Gary, Mahsaeky air Sh 
: ae ec 
But R=Qc+P and R’=Qe+P” (Art. 252). 


Substituting these values in the last equation, we have 
— @ RY (WFP) Q~(QE PG 
\ ray a RQ’ Y 


or reducing 





Om eR PQ PQ’ 


een RQ 
| fae rigs, 
From which we see, that the numerator of the difference Feinyay 
s % ’ 4 
oN > Ou ae 
is equal, with a contrary sign, to that of the difference OR 


That is, the difference between the numerators of any two consecu- 
tive upproximating fractions, when reduced to a common denominator, 
as the same with a contrary sign, as that which ewists between the 
last numerator and the numerator of the fraction immediately fol- 
lowing. 

But we have already seen that the difference of the numerators 
of the Ist and 2d fractions is equal to +1; also that the differ- 
‘ence between the numerators of the 2d and 3d fractions is equal 
‘to —1; hence, the difference between the numerators of the 3d 
and 4th is equal to +1; and so on for the following fractions. 


_ 


IS 


ao SD abbare 


ELEMENTS OF ALGEBRA (CHAP. IX. 


Since the odd approximating fractions are all greater than the 
true value of the continued fraction, and the even ones all less 
(Art. 251), it follows, that when a fraction of an even order is 
subtracted from one of an odd order, the difference should have 
a plus sign; and on the contrary, it ought to have a minus sign, 
when one of an odd order is subtracted from one of an even. 


254. It has already been shown (Art. 251), that each of the 
approximating fractions corresponding to the odd numbers, exceeds 
the true value of the continued fraction; while each of those cor- 
responding to the even numbers, is less than it. Hence, the dif- 
ference between any two consecutive fractions is greater than the 
difference between either of them and the true value of the con- 
tinued fraction. ‘Therefore, stopping at the n” fraction, the result 
will be true to within 1 divided by the denominator of the n™ 
fraction, multiplied by the denominator of the fraction which fol- 
lows.) Thus, if Q’ and R’ are the denominators of consecutive 
fractions, and we stop at the fraction whose denominator is Q’, 





the result will be true to within But since a, 6, c, d, &c., 


1 
Re 


are entire numbers, the denominator R’ will be greater than Q’, 


and we shall have 
1 con ge ps Pe 
ity alae VG, GAB ts 


ee Ln 
hence, if the result be true to within ——, it will certainly be 


Q R? 


true to within less than the larger quantity 
H i 
oF 1D yp: ‘hb 


that is, the approximate result which is obtained, is true to withih 


unity divided by the square of the denominator of the last approx 
mating fruction that ts employed. 


If we take the fraction ia we have 


829 1 


i 


ie 


\PHAP IX.] EXPONENTIAL QUANTITIES. 269 


{ Here, we have in the quotient the whole number 2, which 
wnay either be set aside, and added to the fractional part after its 
value shall have been found, or we may place 1 under it for a 
lenominator, and treat it as an approximating fraction. 


| | Resolution of the Equation a? = b. 


255. An equation of the form 
{ newt ; 
3 called an exponential equation. The object in resolving this 
“quation is, to find the exponent of the power to which it is ne- 


essary to raise a given number a, in order to produce another 
‘iven number 6. 


| Suppose it were required to resolve the equation 
‘i 2* = 64. 
By raising 2 to its different powers, we find that 2° — 64; 


Nence, x= 6 will satisfy the conditions of the equation. 
_ Again, let there be the equation 


| % 3% = 243, in which x= 5. 


a fact, so long as the second member 3 is a perfect power of 
e given number a, it may be obtained by raising a@ to its suc- 
essive powers, commencing at the first. 


Suppose it were required to resolve the equation 
2° = 6. 
' By making «= 2, and x =3, we find 
27—=4 and 2? = 8; 
om which we perceive: that the value of x is comprised between 
and 3. 


jes , 
| Make then, 2=2-+-—, in which a >]. 





Substituting this value in the given equation, it becomes, 


a8 pa 
Pee = 6, On 27 xX, 27 — 6: hence, 

=: 6 3 

27 = — = —_; 

4 2 


id by changing the terms and raising both members to the 
power, 


| Gy 


eo 


270 ELEMENTS OF ALGEBRA. [CHAP. IX. 


To determine x’, make successively a’ = 1 and 2; we find 
9 . 


2 ae 32 
(—} = — ; —) =— > 23 
(i) eal aa (3) =a- 
therefore, a’ is comprised between 1 and 2, 
Lees : 
Make, a =1+ oa in which 2” >1. 


a sUNts : oihwt 
By substituting this value in the equation (=) = 2, 


3 \1+—. 3 3\— 
(=) w=). Hence: is x (=) ==: 2, 
nd consequentl (=)" = : 
a quently, 3 = aa 
; 4 3 
The hypothesis «’=1, gives 7 <t Py 
{ 165)" *98 
and of Hind Peer 6) x xs 
therefore, a” is comprised between 1 and 2. . 
1 f 
Let x’ = 1 -+- “pit then, : 
AN WS 4 4\—-' 3 
(=) GIT ee hence, z x (=)P"= 93 
ante | 
Onert 
whence, 3 3 
If we make 2” = 2, we have 
(ee ogee 
ghee Se 
and if we make 2’ = 3, we have 
( 9 i Le S 
Bele .,'- Silat 
therefore, «’” is comprised between 2 and 3. 
Make we’ — 2+ a and we have 
Ao ea. 
(<) r= nk hence, 81 (=) giv — 5 ; 
8 3 64 \8 F foal 
nd consequent ee = : 
7 aan 2497" one 


! 
i 
} 
| 
: 


AAP. IX.] EXPONENTIAL QUANTITIES. 27 


Operating upon this exponential equation in the same manner 
3 upon the preceding equations, we shall find two entire num- 
ers, & and k+ 1, between which «'Y will be comprised. 
Making 
1 
av okt —, 
a 
nd x’ can be determined in the same manner as z'’, and so on. 
_Making the necessary substitutions in the equations 





2 1 1 1 
me 24+—, weal+—, WH14+— e%=24+ 
x ao 


wv” mv ae 9g 


re obtain the value of « under the form of a continued fraction 
i 


a 





x£—=2+ 








i Hence, we find the first three approximating fractions to be 
quite, B 
TA 9” re 
nd the fourth is equal to 
| 2 te a 
ae chao 


hich is the true value of the fractional part of « to within 


‘3 
— RTs | 
Ta (Ar 2), 


1 
(12)? or hat, (Art. 254). 


Therefore, 





1 

144’ 

ad if a greater degree of exactness is required, we must take a 
reater number of integral fractions. 


? 31 she 
ee 2+ 12 = 12 = 2.58333 + ” within 


EXAMPLES. 
OS 15 ee - oe 2.46 to within 0.01. 
107= 3 - - - «= 0.477. 0.001. 


| a eee ee 0.25 ss 0.01. 


292 ELEMENTS OF ALGEBRA. [CHAP. Ix 


Theory of Logarithms. 


\ 


256. If we suppose a@ to preserve the same value in the equatio: 
ei, 
and NV become, in succession, every possible positive number, | 
is plain that # will undergo changes corresponding to those mad 
in IV. By the method explained in the last Article, we can de 
termine, for each value of N, the corresponding value of x, eithe 
exactly or approximatively. 

Any number, except 1, may be taken for the invariable num 
ber a; but when once chosen, it is supposed to remain the sain 
for the formation of one entire series of numbers. 

The exponent « of a, corresponding to any value of N, is calle 
the logarithm of that number; and the invariable number a is calle 
the base of that system of logarithms. Hence, 


The logarithm of a number, is the exponent of the power to whic 
it is necessary to raise an invariable number, called the base of th 
system, in order to produce the number. 

The general properties of logarithms are independent of am 
particular base. ‘The use that may be made of them in nn 
merical calculations, supposes the construction of a table, con 
taining all the numbers in one column, and the logarithms 0! 
these numbers in another, calculated from a given base. Now, ii 
calculating this table, it is necessary, in considering the equation 

a® = N, 
to make NV pass through all possible states of value, and to de 


termine the value of « corresponding to each of the values of WV 
which may be done by the method of Art. 255. 


257. The base of the common system of logarithms, or as the: 
are sometimes called, Briggs’ logarithms, from their inventor, i: 
the nnmber 10. If we designate the logarithm of any numbe 
by log. or J, we shall have 

CTO)" = 1; hence, | log. Lome 

(10)! 10; hence, . logy, 00am 

(10)? 100; “hence, ‘log, (100i 

(10)? = 1000;' hence, ‘log. 1000 ==3% 

(10)*= 10000; hence, log. 10000 = 4. 
&c., &e. 


II 


| 


y 
i 


t 
MAP. IX.] THEORY OF LOGARITHMS. 273 


Hence, in the common system, the logarithm of any number 
etween 1 and 10, is >0 and <1. The logarithm of any num- 
er between 10 and 100, is >1 and <2; the logarithm of any 
umber between 100 and 1000, is°>2 and <3; and so on. 


Hence, the logarithm of any number expressed by two figures, 
nl which is not a perfect power of the base of the system, 
aul be equal to a whole number plus an approximating fraction, 
te approximate value of which fraction is generally expressed 
ecimally. : 


The integral part of a logarithm, is called the index or char- 
steristic of the logarithm. 

_ By examining the several powers of 10, we see, that if a num- 
er is expressed by a single figure, the characteristic of its logarithm 
ill be 0; if it is expressed by two figures, the characteristic of 
3 logarithm will be 1; if it is expressed by three figures, the 
aaracteristic will be 2; and if it is expressed by n places of 
gures, the characteristic will be » —1 units. 


The following table shows the logarithms of the numbers, from 
} : 
to 100. 





~~ Log. he a Logs soll) N. Log. N.. | Log. .° | 
0.000000 26 1.414973 | 51 1.707570 || 76 _ 1.880814 
0.301030 27 1.431364 || 52 | 1.716003 77 | 1.886491 
0.477121 28 1.447158 || 53 1.724276 || 78 , 1.892095 
0.602060 29 1.462398 ||'54 | 1.732394 || 79 | 1.897627 
0.698970 30 1.477121 || 55.) (1.740863 80 |. 1.903090 | 
0.778151 31 1.491362 56 | 1.748188 8] 1.908485 | 
0.845098 32 1.505150 || 57 | 1.755875 || 82 | 1.913814 ; 
0.903090 33 1.518514 || 58 | 1.763428 83 | 1.919078 , 
0.954243 34 1.531479 59 | 1.770852 84 | 1.924279 
1.000000 35 1.544068 |; 60 | 1.778151 85 | 1.929419 


11 1.041393 36 1.556303 | 1.785330 86 | 1.934498 
12 |. 1.079181 || 37 1.568202 || 62 | 1.792392 87 | 1.930519 | 
13 1.113943 38 1.579784 || 63 | 1.799341 88 | 1.944483 
14 | 1.146128 || 39 1.591065 || 64 | 1.806180 89 | 1.949390 
15 1.176091 40 1.602060 || 65 | 1.812913 90 | 1.954243 


1.204120 1.612784 1.819544 91 | 1.959041 
17 | 1.230449 |} 42 | 1.623249 || 67 | 1.826075 92 | 1.963788 





Se ee er a 








| Scwmral om ow! 


me as ne 





=| 


= 
fon 
aad 
rp) 
[op 

















<n ee 








18 | 1.255273 || 43 | 1.633468 || 68 | 1.832509 || 93 | 1.968483 
19 | 1.278754 || 44 | 1.643453 || 69 | 1.838849 || 94 | 1.973128 
20 | 1.301030 || 45 | 1.653213 || 70 | 1.845098 || 95 | 1.977724 
21 | 1.322219 || 46 | 1.662758 || 71 | 1.851258.|| 96 | 1.982271 
22 | 1.342493 || 47 | 1.672098 || 72 | 1.857333 || 97 | 1.986772 | 
23 | 1.361728 | 48 | 1.681241 || 73 | 1.863323 || 98 | 1.991226 | 
24 | 1.380211 49 | 1.690196 || 74 | 1.869232 || 99 | 1.995635 | 
25 | 1.297940 |, 50 | 1.698970 || 75 | 1.875061 || 100 | 2.000000 | 


274 ELEMENYS OF ALGEBRA. [CHAP. 1X 


The characteristic being always one less than the number o 
places of figures in the number, is not written down in the tabl 
of logarithms for numbers which exceed 100. ‘Thus, in searck 
ing for the logarithm of 2970, we should find in the table oppcsit 
2970, the decimal part .472756. But since the number is e} 
pressed by four figures, the characteristic of the logarithm is ‘ 


Hence, log.2970 = 3.472756, 

and by the definition of a logarithm, the equation 
a” = N, gives 
103-472756 — 9970, 


Multiplication and Division by Logarithms. 


258. Let a@ be the base of a system of logarithms, and sy 
pose the table to be calculated. Let it be required to multip! 
together a series of numbers by means of their logarithms. D 
note the numbers by NV, N’, N”, N’’, &c., and their correspon 
ing logarithms by a, w’, vw, #”, &c. Then, by definition (Ar 
257), we have 

ess N, ea N, a?” = N”, BE re EAS he” GLC: 

Multiplying these equations together, member by member, ar 

applying the rule for the exponents, we have 


gitetel/+al// ee dg en x N x NY” x N” . 
But since ais the base of the system, we have 
wise ia! earl + af. . = log (NYA ee eee i 


that is, the sum of the logarithms of any number of factors, is equ 
to the logarithm of the product of those factors. 


259. Suppose it were required to divide one number by anothe 
Let N and WN’ denote the numbers, and 2 and 2’ their logarithm 
We have the equations | 

a®='N “and “ot tir: 





hence, by division ~ — gt-v oa : 
or z — x = log. N— log. NW’ = log. (=) 


that is, the difference between the logarithm of the dividend and t 
logarithm of the divisor, is equal to the logarithm of the quotient 


AP. IX.] THEORY OF LOGARITHMS. 275 


Consequences of these Properties. 
A multiplication can be performed by taking the logarithms of 
‘two factors from the tables, and adding them together ; this 
T give the logarithm of the product. Then finding this new 
arithm in the tables, and taking the number which corresponds 
it, we shall obtain the required product. Therefore, by a sim- 
addition, we find the product arising from a multiplication. 
in like manner, when one number is to be divided by another, 
itract the logarithm of the divisor from that of the dividend, 
n find the number corresponding to this difference; this will 
‘the required quotient. Therefore, by a simple subtraction, we 
ain the quotient arising from a division. 
j 
Formation of Powers and Extraction of Roots. 
260. Let it be required to raise a number N to any power de- 
} m 
ed by ~ If a denotes the base of the system, and 2 the 
arithm of N, we shall have 

@ =i N, or... N= a*; 


He m 
ence, by raising both members to the power — 
n 


b] 


™m hag 
WW Sia*” 
ey 
as m 
erefore, log. (v7) a= a = log. NV. 


f we make n= 1; there will result, 
m.log. N = log. N™; 
‘equation which is susceptible of the following enunciation : 


f the logarithm of any number be multiplied by the exponent of 
power to which the number is to be raised, the product will be 


al to the logarithm of that power. 


61. Suppose, in the first equation, m= 1; there will result, 
1 ou 
me log. N = log. N* = log. */N; that is, 
i) 


The logarithm of any root of a number is obtained, by diwi hng 
logarithm of the number by the index of the root. 


Ais 


276 ELEMENTS OF ALGEBRA. (CHAP. Ix 


Consequences. 

To form any power of a number, take the sclaaeuln of thi 
number. from the tables, multiply it by the exponent of the power 
then the number corresponding to this product will be the require 
power. 

In like manner, to extract the root of a number, divide the log 
arithm of the proposed number by the index of the root; the 
the number corresponding to the quotient will be the require 
root. Therefore, by a simple multiplication, we can raise a quar 
tity to a power, and extract its root by a simple division. 


262. If we make the different exponents of 10 negative, th 
powers corresponding thereto will be decimal fractions. Thus, 


1 
OL) ie ry ee On; hence, __log.0.1 =—1,; 
1 
wieder) | tt Ht = --2; 
(10) 100 0.01 ; hence, log.0.01 
1 
“3 = ——_ = 0.001; h .0. =—3; 
(10) 55 = 9.001; hence, log. 0.001 3 
1 
—+— —__._ — 0.0001; h .0.0001 = — 4. 
(10 10000 ence, log.0.000 
&c., &c., &c. 


The logarithm of any fraction between one and one tenth, ar 
as four tenths, for example, may be expressed thus, 


log. (=) ce LO & x 4) = log. = + log.4 = — 1 + log. 4. 


For the fractions between one hundredth and one tenth, as si 
hundredths, for example, we have : 


6 1 
log. (—.) = log. (5 x 6) = — 2 + log. 6. 
For the fractions between one thousandth and one hundredt 
as eight thousandths, for example, we have 


log. (35) — log. (— x 8 ) = — 3 + log. 8. 


Now, instead of performing the subtractions indicated above, v 
unite the decimal part of the logarithm to the negative chara 
teristic. Thus, 

log.0.4 =—1-+ log.4 = — 1.602060; 
log.0.06 = —2-+ log.6 = — 2.778151 ; 
log. 0.008 = — 3 + log.8 = — 3.903090. 


HAP. IX.] THEORY OF LOGARITHMS. Q77 


_ Adopting this method of writing the logarithms, we see that the 
ygarithm of a decimal fraction may be found from the tables, by 
mting to the logarithm of its numerator, regarded as a whole num- 
r, a negative characteristic greater by unity than the number of 
iphers between the decimal point and the first significant figure. 
To demonstrate this in a general manner, let a denote the nu- 
ierator of a decimal fraction, and 6 its denominator. From the 
ature of decimals, we shall have 


b= (10)s, 
i which m will denote the number of ciphers in the denomina- 


w. Hence 


a a 
Pe log. ste log. (Gia) = log. a — m log. 10 = log. a — m. 


Or, in other words, the logarithm of a whole number will be- 
me the logarithm of a corresponding decimal, by adding to it a 
egative characteristic containing as many units as there are ciphers 
1 the denominator of the decimal fraction. 

“Hence, the table of logarithms whose base is 10, will give the 
garithms of all decimals, as well as of the integral numbers. 


GENERAL EXAMPLES. 


]. What is the square of 7? 





Miata. 2 Wend sony Sy . = 01845098 
| Exponent of the power - =_ 2 
Number corresponding, 49 - 1.690196. 
2. What is the 6th power of 2? 
Log.2 - - - : - = 0.301030 
| Exponent of the power - - == 6 SR, 
Number corresponding, 64 - 1.806180. se 
ponding | Vo 
3. What is the cube root of 64? 
: Log. 64 + EEE Son \~Soon = 4BOBk8O 
| Then, - - - - - 3) 1.806180 
Number corresponding, 4 - 0.603060. 
4, What is the 4th root of 81? Ans. 3. 


5. What is the 5th root of 32? Ans. 2 


278 ELEMENTS OF ALGEBRA. [CHAP. 1 


6. Log. (a.b.c.d....) =log.a + log.b + log.c.... 


7. Log. fo = log.a + log.b + log. c—log. dhe 
8. Log. \(a™. "cP. ....) = mlog. a + nlog bp plop. c + & 
9. Log. (a?— a?) = log. (a + x) + log. (a — 2). 
10. Log. /(a@ — a?) = 4log.(a+«) + 4log. (a — 2). 
il. Log. (a? x 4/3) = 33 log. a. 
263. Let us resume the general equation 
at = N, 


and suppose a to be the base of a system of logarithms. The 
Ist, we have a! = N=a, whence, log. a=1; 
2d Grd, whence, log. 1=0; 


that is, whatever be the base of the system, its logarithm taken 
that system, is equal to 1, and the logarithm of 1 ts equal to | 


264. Let us suppose, in the equation 


a? == NV, 
that Sipil. 
Then, if we make N=1, we shall have 
a = T, 
If we make N< 1, we must have 
a? sa age <=N<1. 


If now, N diminishes, x will increase, and when NV becomes 
we have ‘ 


a= =-—— =0, or) a= ow (Ari ibe); 


but no finite power of a is infinite, hence «=o: and therefa 
the logarithm of 0 ina system of which the base is greater t! 
unity, is an infinite number and negative. 
265. Again, take, the equation 
a = N, 
and suppose the base a<1. Then making, as before, 
N=1, we have a =—1. 


‘ 


| 
| 


“HAP. IX.] LOGARITHMIC SERIES. 279 


( If we make WN less than 1, we shall have 





aw N ae 


_ Now, if we diminish NV, x will increase; for, since a < 1, its 
yowers will diminish as the exponent « increases, and when 
V=0, « must be infinite, for no finite power of a fraction can 
ve 0. Hence, the logarithm of 0 in a system of which the base 
. less than unity, is an infinite number and positive. 


Logarithmic and Exponential Series. 


_ 266. The method of resolving the equation 
; rane 
Bbininca in Art. 255, gives an idea of the construction of log- 
rithmic tables; but this method is laborious when it is necessary - 
(0 approximate very near the value of x. Analysts have discovered 
auch more expeditious methods for constructing new tables, or for 
verifying those already calculated. These methods consist in the 
‘evelopment of logarithms into series. 
Taking again the equation 

a ty; 
; is proposed to develop the logarithm of y into a series involving 
he powers of y, and co-efficients independent of y. 

It is evident, that the same number y will have a different log- 
Tithm in different systems, that is, for different values of the base 
; hence, the log. y, will depend for its value, 1st, on the value 
f y; and 2dly, on a, the base of the system of logarithms. 
lence, the development must contain y, or some quantity depen- 








ent on it, and some quantity dependent on the base a. 
To find the form of tlis development, we will assume 


log. y= A+ By + Cy? + Dy? 4+, &e., 


1 which A, B, C, &c., are independent of y, and dependent on 
1e base a. 

Now, if we make y = 0, the log. y becomes infinite, and is 
ither negative or positive, according as the base a is greater or 
‘ss than unity (Arts. 264 & 265). But the second member un- 
er this supposition, reduces to A,.a finite number: hence, the 
evelopment cannot be made under that form 


280 ELEMENTS OF ALGEBRA. (CHAP. x. 


Again, assume 
log. y = Ay + By? + Cy? + Dy+ +, &e. 
If we make y = 0, we have 
log..y == + @, that isy 2: wee), 
which is absurd, and hence the development cannot be made un- 
der the last form. Hence we conclude that, the logarithm of a 
number cannot be developed in the powers of that number. 

Let us place, in the first member, 1 + y for y, and we have 
log. (1 + y) = Ay + By? + Cy? + Dyt + &e.... (1), 
making y= 0, the equation is reduced to log. 1 =0, which does 

not present any absurdity. 

In order to determine the co-efficients A, B, C, . .. we shall 
follow the process of Art. 243. Since equation (1) is true for 
any value of y, it will be true if we substitute z for y, and we 
may write 


log. (1 +2) = Az+ B2?+ C2? + DA+ ... (2). 
Subtracting equation (2) from (1), we obtain 
log. (1+y)—log.(1+z) =A(y—z)+ B(y?—27)+ C(y8—z%)+ .. (3), 
The second member of this equation is divisible by y—z. Lei 


us see, if we can by any artifice, put the first under such a fore 
that it shall also be divisible by y — z. We have, 


log.(1 +y) —log.(1 + z)= log. ( 1) = logs(1 + 4} 


Y= 
1 





But since 





can be regarded as a single number wu, wi 





can develop log.(1-+%), or log. ( 





— ee Ae 
, ), in the same mai: 
zZ 


ner as log.(1 + y), which gives 


eG ae ee 4452+ B(t = Y= *)'4.0 (454) 4. 
Substituting this development for 


log. (1 + y) — log. (1 + 2), 
in the equation (3), and dividing both members Py y — 2, it be: 
comes 


J seid (y — 2)? 
Srna! tay + Ole eye 


=A+Biy+2)+C(y+yz+2)+... 
















“CHAP. IX.] LOGARITHMIC SERIES. 281 


_ Since this equation, like the preceding, is true for all values of 
y and z, make y =z, and there will result 


re = A+ 2By + 3Cy? + 4Dy> + 5Ey* + . 


whence, by making the terms entire, and cus. 






































o= 37 Mea) eB) se 30 | 4 aD 


Placing the co-efficients of the diffetent powers of y equal to 
zero, we obtain the series of equations 


me A= 0/ 9R+ A-0, 30+2B=0, 4D4+3C=0...: 
whence, 


} A 2B 3C A 
5 —A Se witers, coc iat onyee T a eng ge EE, he! ee ee 
A hee 2° Sits a 4 4 4 


The law of the series is evident; the co-efficient of the n™ term 
‘ A E : 

fm equal to = —, according as n is even or odd: hence, we ob- 
n 

\tain for the development, 


opel ba) = Ay. — sgt chic ey ty 


2 3 4 5 
ee ane 

Hence, although the logarithm of a number cannot be developed 

in the powers of that number, yet it may be developed in the powers 

of a number less by unity. 

By the above method of development, the co-efficients B, C, D, 

\E, é&c., have all been determined» in functions of A; but A re- 

|Mains entirely undetermined. This indeed should be so, since A 

lepends for its value on the base of the system, to which any 

}ralue may be assigned. 

Denote by 2’ that part of the second member of equation (4) 

which involves y, and suppose a to be the base of the system in 

| jwhich the log.(1 + y) is taken, and we have 


a4t’/—1+y, or Aa’ =log.(1+y). 
But the log. (1 + y) depends for its value on two things: viz a, 


282 ELEMENTS OF ALGEBRA. [CHAP. Ix, 


pressed in y, and hence depends for its value on y alone. But 
A being independent of y, its value must depend on the base of 
the system; and hence, 

The expression for the logarithm of any number is composed of 
two fuctors, one dependent on the number, and the other on the base 
of the system in which the logarithm is taken. The factor which 
depends’ on the base, is called the modulus of the system of log- 
arithms. 


267. If we take the logarithm of 1+ y in a new system, and 
denote it by I’.(1 + y), we shall have 


2 3 4 5 
Vaty=sa (y-S+5-442 m&e.) (5), 


in which A’ is the modulus of the new system. 

If we suppose y to have the same value as in equation (4), we 

shall have 

V.(l+y):Lil+y):: Als ae 
for, since the series in the second members are the same, they 
may be omitted. Therefore, 

The logarithms of the same number, taken in two different systems, 
are to each other as the moduli of those systems. 

268. Having shown that the modulus and base of a system of 
logarithms are mutually dependent on each other, it follows, that 
if a value be assigned to one of them, the corresponding value of 
the other must be determined from it. 

If then, we make the modulus 

are). 
the base of the system will assumo 2 fixed value. The system 
of logarithms resulting from such a modus, u..i such a base, 1s 
called the Naperian System. This was the first system known, 
and was invented by Baron Napier, a Scotch mathematician. 

With this modification, the proportion above becomes 

BH +9) SCD 9) 2a ee 
and A.l.(1 + y) =1.(1 + y). 
Hence we see that, 
The Naperian logarithm of any number, multiplied by the mody- 


lus of another system, will give the logarithm of the same numbet 
en that system. 





‘ CHAP. IX.] LOGARITHMIC SERIES. 283 


. The modulus of the Naperian System being unity, it is found 
most convenient to compare all other systems with the Naperian; 
and hence, the modulus of any system may be defined to be, 
| The number by which it is necessary to multiply the Naperian 
logarithm in order to obtain the logarithm of the same number in 

the other system. 


269. Again, AxlV.ali+y)=1l(1+y) gives 


That is, the logarithm of any number divided by the modulus of 
its system, ts equal to the Naperian logarithm of the same number. 


270. If-we take the Naperian logarithm and make y = 1, equa- 
tion (5) becomes 


1 1 1 miss 
Y.2=1——+———4-—... 
77 3 asa iG 


‘a series which does not converge rapidly, and in which it would 
| be necessary to take a great number of terms to obtain a near ap- 
proximation. In general, this series will not serve for determining 
the logarithms of entire numbers, since for every number greater 
than 2 we should obtain a series in which the terms would go 
on increasing continually. 

The following are the principal transformations for converting 
the above series into converging series, for the purpose of obtain- 
ing the logarithms of entire numbers, which are the only logarithms 
placed in the tables. 


First Transformation. 


i 
Taking the Naperian logarithm in equation (5), making y=— 


and observing that 
1 
/ ary eros saris : 
v.(1 a5 —) =l.(1+2)—IV.z, it becomes 


| | 1 1 1 1 

V.di+2)—V.2=— — ——- + ~~ — —— + &e. (6). 

pi thee) Zz 227 323 424 v (6) 
This series becomes more converging as z increases; besides, 

tne first member of the equation expresses the difference between 

the logarithms of two consecutive numbers. 


\ 


284 ELEMENTS OF ALGEBRA. (CHAP. IX. 


Making z= 1, 2, 3, 4, 5, &c., in succession, we have 


V2=1-— +>-T+7— be 
V.3—V2=—— + apege ae 
1 
‘ V¥4—-V3 => — =t+o-urt ra 
1 
V.5—-V4=2— 24 


The first series will give the logarithm of 2; the second series 
will give the logarithm of 3 by means of the logarithm of 2; the 
third, the logarithm of 4, in functions of the logarithm of 3... &c. 
The degree of approximation can be estimated, since the series 
are composed of terms alternately positive and negative (Art. 241). 


Second Transformation. 


A much more converging series is obtained in the following 
manner. In the series 
yO ae 
eT @! =xr—-—+————4... 
(Lith #) ee oo Suche 


substitute — x for x, and it becomes 
Vl —2) = —a# —-— —- — — — — Be 95: 


Subtracting the second series from the first, observing that 


l+e2 


iar ), we obtain 
x 





V.(1+2)—-V.0 -2) =V.(+ 





a es 


1—~we 


ee eee 
)a2(e+ 5+ e+ Gt Gt.) 
This series will not converge very rapidly unless 2 is a very 


: : . = eS] 
small fraction, in which case, — will be greater than unity, 





but will differ very little from it. ‘ 
l+e 1 ; ; 
Make “DRS 1+ Par being an entire number. We have 


1 
(1+ 2)2=(1 —2)(z+1); whence, & aah 


| HAP. IX.] LOGARITHMIC SERIES. 285 
; 1 
| Hence, the preceding series becomes I’. (1 + —), or 


| i | 1 
Meri veteiue ye wid Na Po: rath ) 
( 22-1 ' 3Q2+1) Sepp): 
This series gives the difference between the logarithms of two 
|omsecutive numbers, and converges more rapidly than series (6) 
| Making successively, z= 1, 2, 3, 4, 5..., we find 
| ene een tee 
Be ites emt Oy ye yh 
1 ] 1 1 


3 + 3.53 5.58 + 7p 













y2s2( + .. 


V.3—V2—2 swe: 











ee © © @o fe 


) 

( ) 
ramet ys Assy 
( 


\ Let z= 100; there will result 


Asie 1 1 l 
V.101 =V.100+4+2(5— + Rcaa's o0ry 
| thence we see, that knowing the logarithm of 100, the first term 
|f the series is sufficient for obtaining that of 101 to seven places 
_f decimals. 

| There are formulas more converging than the above, from which 
re may obtain a series of logarithms in functions of others al- 
| sady known, but the preceding are sufficient to give an idea of 
ine facility with which tables may be constructed. We may now 
|uppose the Naperian logarithms’ of all numbers to be known. 
| The Naperian logarithm of 10 may be deduced from the first 
pnd fourth of the above equations, by simply adding the logarithm 
)f£ 2 to that of 5 (Art. 258). This number has been calculated 
‘ith great exactness, and is 2.302585093. 


+...); 


| 271. We have already observed that the base of the common 
tystem of logarithms is 10 (Art. 257). We will now find its 


i 


jrodulus. We have, 
V(l+y):L(i+y)::1: A (Art. 267). 


L 


286 ELEMENTS OF ALGEBRA. [CHAP. Ix, 


If we make y = 9, we shall have 
VADs: 1. 1002S es eee 


But the 1.10 = 2.302585093, and 1.10 =1 (Art. 257) ; 
1 
hence, A = 950555009 0484294482 == the modulus of the 
common system. 

If now, we multiply the Naperian logarithms before found, by 
this modulus, we shall obtain a table of common logarithms 
(Art. 268). 

All that now remains to be done is to find the base of the Na- 
perian system. If we designate that base by e, we shall have 
(Art. 267), 

Ve: le: : 1: 0.4384294482. 


But Ve = 1 (srt.' 263): hence, 
l.: lLe:: 1: 0.434294482, 
hence, lye = 0.434294482. 


But as we have already explained the method: of calculating 
the common tables, we may use them to find the number whose 
logarithm is 0.434294482, which we shall find to be 2.718281828: 
hence 

e = 2.718281828. 

We see from the last equation but one that, the modulus of the 

common system is equal to the common logarithm of the Naperian base 


Of Interpolation. 


272. A table of logarithms is a tabulated series of numbers, 

showing the value of x in the equation 

a = N, 
corresponding to all the integral values of NV, between 1 and some 
higher number which marks the limit of the table. It has already 
been remarked that in the system in common use, the value of the 
base a, is 10. 

And generally, any mathematical table consists of a series of values 
of some letter in an algebraic expression, corresponding to equi-dis 
tant values of the function on which it depends. 

The principle of interpolation, which is of great value in prac 
tical science, has for its object to find from the tabulated numbers 











(AP. IX] oF INTERPOLATION. 287 


hich are given, other similar numbers which shall correspond 
, intermediate values of the function. For example, suppose p, g, 
s, &c., to be a series of tabulated numbers corresponding to, 
id written opposite the functions a, a+b, a+ 2b, a+ 36, &c., 
‘id it were required to find the tabulated number corresponding 
the function a +236. This is a question of interpolation, and 
resolved by taking the successive differences of the tabulated 
‘umbers, thus: 


Functions. | Differences of tabulated numbers 
a p - - - 
dp - - 
‘ a+ 6b| q dp - 
dq a3 
a+ 20) r d’q 
dr d°q 
a+ 3b)| s d’y,  - 
. ds - - 
| a+ 4b | t Te 
oP te 4 x 
‘1. which dp = 4 —p, dg=r—q, dr=s—r, &c; 


ad dp =—dg—dp, @g=dr—dg, dr=ds —dr, &c.; 
dso, Bp =dg—dp, dg= dr — dq, &e. 
&c., &c. 
From the above equations, we have 
gq=pt+dp, r=qtdg=p+dp+dq=p-+ 2dp + dp; 
nd by a similar process, we have 
s= p+ 3dp + 3d2p + dp, 
t= p+ 4dp + 6d?p + 4d%p + d*p, 
; &c., &c., 
‘) which notation it should be observed, that d?, d3, &c., denote the 
econd, third, &c. differences of the successive tabulated numbers 
It is plain, that the above law from which the numerical co- 
ficients for any term, may be derived, is similar to that for the 
o-efficients of a binomial: hence, if JT’ denote the n+1 term of 
ae tabulated numbers, reckoning from p inclusive, we shall have 


{ 


Li horme eam, er 1) ee) 
Os ui aa ei ee a oP 


2838 ELEMENTS OF ALGEBRA. . [CHAP. 1x 


Let it be required to find the tabulated number correspondin, 
to a+ 3b. We then have, n= 3: hence, 


T = p + 3dp + 3d’p + dp, 


the same value as that found above for r. 


Next, let it be required to find the tabulated value answerin; 
to the functions a + 36. Then, n= 3%, and if we know the tab 
ulated number p, and the successive differences d, d?, &c., th 


approximate value of T' can easily be found. 


It is plain from the series that the interpolated values are bu 
approximations, since no order of difference can reduce to zero 
and hence, the series will contain an infinite number of terms 
Generally, however, the tabulated values are themselves but ap 
proximations, and the successive differences decrease so rapidh 
in value, that the series becomes very converging. 


Let us suppose for example, that we have the logarithms o! 
12, 13, 14, 15, &c., and that it is required to find the logarithn 
of 12 and a half. Then, 


12 , 1.079181 
| dp = 0.034762 | 
13 | 1.113943 @p — — 0.002577 
dg = 0.032185 
14 | 1.146128 d?g = — 0.002222 
dr = 0.029963 
15 | 1.176091 
also, d>p = d?q — d*p = + 0.000355. 
Making n = = and stopping at the term involving the thir 


difference, we have 
j 


1 1 ] 1 1 j 
ae at Sayed by eee BEN PRS a pe ; 
Piri dpi qipts* x a oP + ke 


4 

fileoy ol wily ete. ih 
Lt dy OL PUN) > 
+ i’p - - - - = = 0.000322 
Bi sm el, = 


\ —___ —___—— 


T = log.12} = 1.096906. 


: 
HAP. IX.] FORMULAS FOR INTEREST. 289 





‘ INTEREST. 


273. The solution of all questions relating to interest, may be 
-reatly simplified by employing the algebraic formulas. 
_ In treating of this subject, we shall employ the following no- 
jation : 


Let p= the amount bearing interest, called the principal ; 


; r= the part of $1, which expresses its interest for one 
year, called the rate per cent. ; 

| t= the time that p draws interest ; 

a= the interest of p dollars for ¢ years; 


S=p+ the interest which accrues in the time ¢, which . 
is called the amount. 


! Sunple Interest. 


| To find the interest of a sum p for t years, at the rate r, and 
‘he amount then due. 

Since r denotes the part of a dollar which expresses its in- 
erest for a single year, the interest of p dollars for the same time 
vill be expressed by pr; and for ¢ years it will be ¢ times as 
‘ouch: hence, 

Ligeia A Tae (is 

nd for the amount due, 


| | S=p+pir=p(1+ir).. (2). 


EXAMPLES. 


1. What is the interest, and what the amount of $365 for three 
‘ears and a half, at the rate of 4 per cent. per annum. Here, 
4 


= s09 neo 


\ 





S = 365 + 51,10 = $416,10. 


| 
t= 3.5; 
i= ptr = 365 x 3.5 x 0.04= $51,10: 
lence, 
19 


290 ELEMENTS OF ALGEBRA. [CHAP. Ix 


Present Value and ‘Discount at Simple Interest. 


The present value of any sum S, due ¢ years hence, is the prin 
cipal p, which put at interest for the time ¢, will produce the 
amount S. 

The discount on any sum due ¢ years hence, is the difference 
between that sum and the present value. 

Lo find the present value of a sum of dollars denoted by S, du 
t years hence, at simple interest, at the rate r; also, the discount. 

We have, from formula (2) 


S =p + pir; 


and since p is the principal which in ¢ years will produce the 
sum S, we have 


and for the discount, which we will denote by D, we have 

S Str 
iw > eee (4). 

1. Required the discount on $100, due 3 months hence, at the 
rate of 54 per cent. per annum. 


D=S— 








S = $100 = $100 
t= 3 months = 0.25 
5.5 
on == 0.0. 
Hence, the present value p is 
S 100 ; 
hence, D = S — p= 100 — 98,643 = $1,35,7. 


Compound Interest. 


Compound interest is when the interest on a sum of money be- 
coming due, and not paid, is added to the principal, and the in- 
terest then calculated on this amount as on a new principal. 

To find the amount of a sum p placed at interest for t years, 
compound interest being allowed annually at the rate r. 

At the end of one year the amount will be 


S = \p.-+ pr= p(l 4). 


y 
AP. IX.) FORMULAS FOR INTEREST. 291 


Since compound interest is allowed, this sum now becomes the 
‘incipal, and hence at the end of the second year the amount 
‘ill be 
) S’=p(1+r)+pr(l+r)=p(i+r). 

Regard p(1-+ 7)? as a new principal; we have, at the end of 
‘e third year, | 


S’=p(l+r)?+pr(1+rP=p(1 +r); 

id at the end of ¢ years, 
| ay (14 FORT. Jette (5). 

And from Article 260 we have 
log. S = log. p + tlog. (1 + 7); 

id if any three of the four quantities S, p, t, and r, are given, 
e remaining one can be determined. 

Let it be required to find the time in which a sum p will double 


self at compound interest, the rate being 4 per cent. per annum. 
We have from equation (5), 
(im. 


| S =p (1+ 1) 
But by the conditions of the question, 
| S = 2p = p(1 +r): 
ance, 2=(1-+ 7), 


_ _log.2 ~—_—(0.301030 
~ log.(1 +r) 0.017033 
= 17.673 years 


id 


= 17 years, 8 months, 2 days. 


To find the Discount. 


‘The discount being the difference between the sum S and p, 
e have 


S 1 
=~ ary = 50a 


29:2 ELEMENTS OF ALGEBRA |CHAP, X 


CHAPTER X. 


GENERAL THEORY OF EQUATIONS. 


~ 


274. Tue most celebrated analysts have tried to resolve equé 
tions of any degree whatever, but hitherto their efforts have bee 
unsuccessful with respect to equations of a higher degree than th 
fourth. However, their investigations have conducted them to som 
properties common to equations of every degree, which they hay 
since used, either to resolve certain classes of equations, or to r 
duce the resolution of a given equation to that of one more simpl 
In this chapter it is proposed to make known these propertie 
and their use in facilitating the resolution of equations. 

The development of the properties of equations of any degre 
leads to the consideration of polynomials of a particular natur 
and entirely different from those considered in the first chapte 
These are expressions of the form 


Aa™ + Ba™1+ Caom2?2+ ... + Tx U, 
in which m is a positive whole number; but the co-efficien 
A, B, C,... T, U, any quantities whatever, that is, entire ¢ 


fractional, commensurable or incommensurable. Now, in algebra’ 
division, as explained in Chapter II., the object was this, viz. 
having given two polynomials, entire with reference to all the le 
ters and particular numbers involved in them, ¢o find a third pol 
nomial of the same kind, which multiplied by the second shall pr 
duce the first. 

But when we have two ‘polynomials, 


Ax™ + Br + Cam? + ...4 Tx + JU, 

A’g® + Blyml | Clam? + 0 ee 
which are necessarily entire only she respect to a, and in whic 
the co-efficients A, B, C..., A’, B’, C’..., are any qual 
tities whatever, it may be ee ie find a third polynomial, ¢ 


q 


I 
AP. X.] GENERAL THEORY OF EQUATIONS. 293 


e same form and nature as those that are given, which multiplied 
1 the second will re-produce the first. 


275. Ordinary polynomials, that is, polynomials which are en- 
re with reference to all the exponents and co-efficients, are called 
tional and entire polynomials. Polynomials which are only en- 
re with reference to the letter 2, and whose co-efficients are any 
jantities whatever, are called entire functions of x. 


276. Every complete equation of the m” degree, m being a pos- 
ive whole number, may, by the transposition of terms, and by 
te division of both members by the co-efficient of a”, be put un- 
or the form 


. a” + Pom 4 Qom2 4 ...4 Tr +U=0, 


'Q,R..., T, U, being co-efficients taken in the most general 
lgebraic sense. 
Any expression, which substituted in place of x satisfies the equa- 


on, that is, renders its first member equal to 0, is culled a root of 
ve equation. 


277. As every equation may be considered as the translation 
ito algebraic language of the relations which exist between the 
iven and unknown quantities of a problem, we are naturally led 
) suppose that, EVERY EQUATION has at least one root. We will 
Imit this principle, which we shall have occasion to verify here- 
fter for most equations. 

We will now demonstrate some of the principal properties of 
general equation. 

| 


First Property. 


278. In every general equation of the form 
am + Pem-1+ Qam2+ ... 4747+ 0=0, 


te first member is divisible by the difference between the un- 
nown quantity x and a root of the equation; that is, 

Ifa isa root of the equation, the first member will be exactly 
iwistble by x —a; and reciprocally, if a divisor of the form x —a 
wll exactly divide the first member, a will be a root of the equation. 
_ Let us suppose the first member of the proposed equation to be 
ivided by x — a, and the operation continued until all the terms 


294 ELEMENTS OF ALGEBRA [CHAP. : 


involving @ are exhausted: the remainder, if there be any, wi 
then be independent of x. , 

If we represent the remainder by R, and the quotient obtain 
hy Q’, we may write 


om + Peml.... + Te + UH QV (4—a)+R. 


Now, since by hypothesis, a is a root of the equation, if w 
substitute a for x, the first member of the equation will reduce 
zero; the term Q’(a—a) will also reduce to 0, and consequentl 
we shall have 

fi 3h 

But since R does not contain a, its value will not be affect 
by attributing to «2 the particular value a: hence,, the-remainder 
was originally equal to zero, and consequently, the first memb: 
of the equation 

gm + Pom-l it Qyn2.,...4+Tr+U=0, ° 
is exactly divisible by # — a. 

Reciprocally, if a —a is an exact divisor of. the first memb 
of the equation, the quotient Q’ will be exact, and we shall ha 
R=0: hence, 


a+ Pam ...4+Tr+ U= QV (a— a). 


If now, we suppose x= a, the second member will reduce 
zero, consequently, the first will reduce to zero, and hence a w 
be a root of the equation (Art. 276). It is evident, from the n 
ture of division, that the quotient Q’ will be of the form 


Ae aa ai Ts +R’+W=0. 


279. It follows from what has preceded, that in order to di 
cover whether any polynomial is exactly divisible by the binomi 
x —a, it is sufficient to see if the substitution of @ for x w 
reduce the polynomial to zero. 

Reciprocally, if any polynomial is exactly divisible by 2 — 
then we know, that if the polynomial be plaged equal to zero, 
will be a root of the equation. 

The property which we have demonstrated above, enables | 
to diminish the degree of an equation by unity when we kno 
one of its roots, by a simple division; and if two or more of t) 
roots are known, the degree of the equation may be still furth 
diminished by continuing the division. 


rg — a rey 


‘OHAP. X.] GENERAL THEORY OF EQUATIONS. 295 


( EXAMPLES. 
1. One of the roots of the equation 
| xt — 25a? + 602 — 36 = 0 
is 3: what is the equation containing the other roots? 
at — 25x? + 60x — 36 |e — 3 — 3 








a —. 3x3 oe? + 322 — 16x + 12 
: + i 83a3 — 25a? 
t 323 — Qa? 
| 4b 16a? + 60a 
— 16x? — 48x 
12x — 36 
12a — 36 








| Ans. 23 + 3x22 — 16a-+ 12 =0. 
2. Two roots of the equation 
a* — 122? + 482? — 682 + 15 = 0 


are 3 and 5: what is the equation containing the other two? 
Ans. «2 — 4x +1=0. 


3. One of the roots of the equation 
eo? — 6a? + lla —6=0 


is 1: what is the equation containing the other roots? 
Ans. a? — 5e4+6= 0. 


' 
_ 4. Two of the roots of the equation 
4at — 1443 — 5x? + 3la + 6=—0 


are 2 and 3: find the equation containing the other roots. 
Ans. 4x2 + 64+1=0. 


Second Property. 


ee —t 


280. Every equation involving but one unknown quantity, has as 
many roots as there are units in the exponent which denotes its de- 
gree, and no more. 
: Let the proposéd equation be 
" am + Pym) + Qom-2 4+ ...4 Tr + U0 =0. 
. Since every equation is Based t to have at least one root (Art. 
| on if we denote that root by a, the first member will be divisi- 
‘ble by «—a, and we shall have the equation 


ent Pel 4. | = (29 — a) (2-1 + Prom? +...) (1) 





296 ELEMENTS OF ALGEBRA. ~ [CHAP. x 
But if we place | 
—] 4 = ae 


we obtain an equation which has at least one root. 
Denote this root by 6, we have (Art. 278), 


eet ote Ee Se yee a ee b) (2-2 4 PYyr-3 4, ). 


Substituting the second member, for its value in equation (1) 
and we have, 


x™ + Pom-ls = (x — a) (w — b) (a2 4 PY gm-3 4 sas)... Ce 
Reasoning upon the polynomial 
pm 2 4. Digg 3 te 
as upon the preceding polynomial, we have 


si ti i ah 5 di was aah. k c) (a3 4 PYym—4 4 “ik 


and by substitution, 
ae Pam +. == (wa) (8) (a= cee eee 


281. Observe, that for each binomial factor of the first degree with 
reference to a, the degree of x in the polynomial is diminished 
by unity; therefore, after m— 2 factors of the first degree have 
been divided out, the exponent of x will be reduced to m — (m — 2) 
= 2; that is, we shall obtain a polynomial of the second degree 
with reference to 2, which can be decomposed into two factors 
of the first degree (Art. 142), of the form x—k, a —1. Now, 
supposing the m— 2 factors of the first degree to have already 
been indicated, we shall have the identical equation, 


mm Pam + ... = (a — a) (x —b)(e—c)...(~—k) (@ —l) = 0; 


from which we see; that the first member of the proposed equation 

may be decomposed into m binomial factors of the first degree. 
As there is @ root corresponding to each binomial divisor of 

the first degree (Art. 278), it follows that the m binomial factors 


of the first degree, «x —a, x—b, m—e.. +, give the m roots, 
a, b,c..., of the proposed equation. 

But the equation can have no other roots than ee | 
For, if it had a root a’, different from a,b,c... l, it would have 


a divisor «—a’, different from x — a, 2 = b°¢—~e ae ae 
which is impossible. Therefore, finally, 


¢HAP. X.] COMPOSITION OF EQUATIONS. 297 


_ Every equation of the m™ degree has m roots, and can have no 
nore. 





282. In equations which arise from the multiplication of equal 
actors, such as 


(a — a)* (a — 5)? (w — c)? (wx — d) = 0, 


he number of roots is apparently less than the number of units in 
‘he exponent which denotes the degree of the equation. But this 
3 not really so; for, the above ‘equation actually has ten roots, 
our of which are equal to a, three to b, two to c, and one to d. 





It is evident that no quantity a’, different from a, 6, c, d, can 
erify the equation; for, if it had a root a’, the, first member 
ould be divisible by »# — a’, which is impossible. 


Consequence of the second Property. 


283. It has been shown that the first member of every equation 
f the m” degree, has m binomial divisors of the first degree, of 


ne form 


x—a, xw—b, w—c,...ux—k, «—l. 







_If we multiply these divisors together, two and two, three and 
‘bree, &c., we shall obtain as many divisors of the second, third, 
‘ve. degree, with reference to x, as we can form different com- 
‘imations of m quantities, taken two and two, three and three, &c. 
Yow the number of these combinations is expressed by 
m—1 m—-l m—2 
Oar oe - 





te CATE. Os 


Hence, the proposed equation has 
m— | 
m . —~— 

2 


ivisors of the second degree ; 


m—1l m—2 


m. 3 . 3 





ivisors of the third degree ; 


m — 1 m—2 m—3 


AG a RTE 


ivisors of the fourth degree; and so on. 


298 ELEMENTS OF ALGEBRA. [CHAP. X. 


Composition of Equations. 
284. If in the identical equation 
am + Pym1 4+ .,, =(x—a) (a —b)(a@—c)... (x —J), 


we perform the multiplication of four factors in the second mem: 
ber, we have, 


was ans 8 x? — abc |x + abcd 


* 





—O) + ac — abd 
Bei habe — acd < e 
—d + be — bed é 
+ bd 
+ cd 





If we perform the multiplication of the m factors of the second 
inember, and compare the terms of the two members, we shall 
find the following relations between the co-efficients P, Q, R, ... 


T, U, and the roots a, b,c,... 4,1, of the proposed equation, viz., 
—a—b—c... —k—l=P, or at+b+e+...+h44+l=—P, 
ao ae oO ek, PERE LA sas Oe 
—abe—abd... —itkl=R, or abe+abd..... + ikl = — RK: 


+abed...kl=U,'or abed... kl =U. 


The double sign has been placed in the last relation, because 
the product —a x —bx —c... x —Jl will be plus or minus 
according as the degree of the equation is even or odd. Hence 


Ist. The algebraic sum of the roots, taken with contrary signs. 
is equal to the co-efficient of the second term; or, the algebraic 
sum of the roots themselves, is equal to the co-efficient of the 
second term taken with a contrary sign. 

2d. The sum of the products of the roots taken two and two, 
with their respective signs, is equal to the co-efficient of the third 
term. \ 

3d. The sum of the products of the roots taken three and three, 
with their signs changed, is equal to the co-efficient of the fourth 
term; or the co-efficieut of the fourth term, taken with a contrary 
sign, is equal to the sum of the products of the roots taken three 
and three; and so on. 







CHAP. X.] COMPOSITION OF EQUATIONS. 299 


4th. The product of all the roots, is equal to the last term; 
| that is, the product of all the roots, taken with their respective 
signs, is equal to the last term of the equation, taken with its sign, 
when the equation is of an even degree, and with a contrary sign, 
| when the equation is of an odd degree. If one of the roots is equal 
_ to 0, the absolute term will be 0. — 

The properties demonstrated (Art. 143), with respect to equa 
tions of the second degree, are only particular cases of the above. 





Consequences ais 


1. If the co-efficient of the second term of an equation is equal 
| to zero, the term will not appear in the equation; and the sum 
of the positive roots is equal to the sum of the negative roots. 


2. Every commensurable root of an equation is a divisor of the 
last or absolute term. 


EXAMPLES IN THE FORMATION OF EQUATIONS. 
1. Form the equation whose roots are 2, 3, 5, and — 6. 
We have, by simply indicating the multiplication of the factors, 
(a — 2) (x — 3) (4 — 5) (a+ 6) = O. 
But the process may be shortened by detaching the co-efficients 
thus: 


; Bieta) | ade 
ott OF 

Gage ety: Ole 

— § + 25 — 30 


1—10+31 —30 |+6 
6 — 60 + 186 — 180 
Pd 994-156 180, 








Hence, the required equation is 
at — 4x3 — 29x? + 1567 — 180 = 0. 
2. What is the equation whose roots are 1, 2, and — 3? 
Ans. & — 7x +6= 0. 
3. What is the equation whose roots are 3, — 4, 2+ / 3, 
end. 2—4/ 3? Ans. xt — 323 — 15a? + 492 — 12 = 0. 


4, What is the equation whose roots are 3+ 5, 3— vy 5, 
and — 6? Ans. «3? — 32a + 24=0. 


300 ELEMENTS OF ALGEBRA. [CHAP. X. 


5. What is the equation whose roots are 1, —2, 3, —4, 5, 
and — 6? 
Ans. «& + 3x5 — 41at — 87x + 4002? + 4442 — 720 = 0. 


6. What is the equation whose roots are....2+~/Y7—1, 
etic fan 1, and — 3? Ans. 23 — a?wabe Ze ++ 15 = 0. 


Of the greatest Common Divsor. 


285. The greatest common divisor of two polynomials is the 
greatest polynomial, with reference to its exponents and co-efhi- 
cients, that will exactly divide the proposed polynomials. 

If two polynomials be divided by their greatest common divisor, 
the quotients will be prime with respect to each other; that is, they 
will no longer contain a common factor. Hence, 


Two polynomials are prime with respect to each other when they 
have not a common factor. 


Let A and B be two polynomials, D their greatest common 
divisor, and A’, B’, the quotients after division. Then 


A \ ? eB mt 
7 mnie and pa: 


and consequently, 
A= A’x D, and |(B=Pixe 


Now, if A’ and B’ have a common factor d, then d x D would 
be a common divisor of the two polynomials and greater than D, 
either with respect to the exponents or the co-efficients, which 
would be contrary to. the supposition. 

Again, since D exactly divides A and B, every factor of D will 
have a corresponding factor in both A and B. Hence, 


Ist. The greatest common divisor of two polynomials contains as 
factors, all the prime factors common to the two polynomials, 
and does not contain any others. 


286. We will now show that the greatest common divisor of 
two polynomials will divide their remainder after they have been 
divided by each other. 

Let A and B be two polynomials, D their greatest common 
divisor, and suppose A to contain the highest exponent of the let- 
ter with reference to which the polynomials A and B are arranged. 


| 


i X.] GREATEST COMMON DIVISOR. 301 
Having divided A by B, suppose we have a quotient Q and a 
Biainder R. We may then write 


4 ae BR OUR: 


If now, we divide both members of the equation by D, we 
have 


5 = a x Q+ si 

‘and since we suppose A to be divisible by D, the first member of 
the equation will be entire, and consequently, the second member 
‘must also be entire, since an entire quantity cannot be equal to 
a fraction. But since D also divides B, the first term of the sec- 
ond member is entire, and consequently, the second term is also 
entire, and therefore, R is exactly divisible by D. 

We will now show that if D will exactly divide B and R, that 


it will also divide A. For, having divided A by B, as before 
we have 


p A=BxQ+R, and by dividing by D, we obtain 
) A B R 
D si mad Mie _ 

But since we suppose B and R to be divisible by D, and know 
2 to be an entire quantity, the second member of the equation is 


mtire: hence, the first member is also entire, that is, A is ex- 


ictly divisible by D. Hence, 


| 

' Qdly. The greatest common divisor of two polynomials, is the same 
ts that which exists between the least polynomial and their remainder 
ifter division. 


~ 


Remarx.—If either of the polynomials A or B have a factor A’ 
ommon to all its terms, but not common to the other polynomial, 
he common divisor will be found in that part of the polynomial 
| a is multiplied by the factor A’. 


" 287. From: these principles, we have, for finding the greatest 
jommon divisor of two polynomials, the following 





RULE. 


I. Take the first polynomial and suppress all the monomial factors 
ymmon to each of its terms. Do the same with the second polyno- 
vial, and if the factors so suppressed have a common divisor, set tt 
side” as forming a part of the common divisor sought. 





302 ELEMENTS OF ALGEBRA. [CHAP. xX 


II. Having done this, prepare the dividend in such a manner that 
ats first term shall be divisible by the first term of the divisor; then 
perform the division, and suppress in the remainder all the factors 
that are common to the co-efficients of the principal letter. Then 
take this remainder as a divisor, and the second polynomial as a 
dividend, and continue the operation with these polynomials, in the 
same manner as with the preceding. 

III. Continue this series of operations until a remainder 1s ob- 
tained which will exactly divide the preceding divisor: this last divisor 
will be the greatest common divisor; but if a remainder 1s obtained 
which is independent of the principal letter, and which will not divide 
the co-efficients of each of the proposed polynomials, it shows that 
the proposed polynomials are prime with respect to each other, or 
that they have not a common factor. 


EXAMPLES. 


1. Find the greatest common divisor of the polynomials 
a? — a2b + 3ab? — 3b3, and a? — 5ab-+ 40?. 











First Operation. Second Operation. 
a —wb+ 3ab2 — 353 ei Saba a? — 5ab + 40% |e b 
db -8, abt Saha 4b | = 4ab 483 | a — 4b 
Ist rem. 19ad?— 1953 0. 


or, 196? (a — b). 


Hence, a — b is the greatest common divisor. | 
We begin by dividing the polynomial of the highest degree by 
that of the lowest; the quotient is, as we see in the above table, 
a+ 4b, and the remainder 19ah? — 190. | 


But, 19ab? — 1953 = 196? (a — B). 
Now, the factor 1962, will divide this remainder without dividing 
a? — Bab + 407: | 
hence, the factor must be suppressed, and the question is re- 
duced to finding the greatest common divisor between 


a*— 5ab+ 4b? and a—b. 


Dividing the first of these two polynomials by the second, there 
is an exact quotient, a—4b; hence, a—b is the greatest com- 
mon divisor of the two given polynomials. ‘To verify this, let each : 
be divided by a — 8. , 
| 


| 





HAP. X.) GREATEST COMMON DIVISOR. 303 







| 2. Find the greatest common divisor of the polynomials 
3a5 — 5a3b2 + 2ab+ and 2at — 3a2h2 + 4, 
“We first suppress a, which is a factor of each term of the first 
dlynomial: we then have 
3a* — 5a?b? + 20+ || 2at — 3a2b? + Jt. 

We now find that the first term of the dividend will not con- 
in the first term of the divisor. We therefore multiply the divi- 
ond by 2, which merely introduces into the dividend a factor not 
‘mmon to the divisor, and hence does not affect the common divi- 
or sought. We then have 
6at — 10a?b? + 454 || 224 — 3a*h? + b4 
Gat — 9a°b? + 358 gi 
eee eens | pe 

— 6? (a? — $?), 
-We find after division, the remainder — a)? + 64, which we 
ut under the form — 6? (a? — 42). We then suppress — 82, and 
ivide 














2a* — 3a2b? + 54 || a? — 2 
2a* — 2a2b?2 2a? — bh? 
— arb? + bt 
— ab? + 54, 





Hence, a? — 6? is the greatest common divisor. 


3. Let it be required to find the greatest common divisor be- 
veen the two polynomials 


— 363 + 3ab? — ab + a3, and 46? — 5ad + a2, 


rurst Operation. 











_ 1268 +. 1208? = 40% + 42 | 402 — 5ab + a 
strem. - - — 3ab?— a@’hb+ 4a3; — 3b, — 3a 
— 12ab? — 4a2b + 1623 
lrem. - - — 19a) + 1903 f 
7 19a? (— b + a). 
Second Operation. 
462 — 5ab + a’? az b+a 
— abta®?| —4b+a 
0. 


Hence, — 6+ <a, or a — 4, is the greatest common divisor. 


304 ELEMENTS OF ALGEBRA. [CHAP. x 


In the first operation we meet with a difficulty in dividing the 
two polynomials, because the first term of the dividend is not 
exactly divisible by the first term of the divisor. But if we ob. 
serve that the co-efficient 4, is not a factor of all the terms of 
the polynomial 


4b? — 5ab + a?, 


and therefore, by the first principle, that 4 cannot form a part oj 
the greatest common divisor, we can, without affecting this com 
mon divisor, introduce this factor into the dividend. This gives 


— 1253 + 12ab? — 4a + 4a’, 


and then the division of the terms is possible. 
Effecting this division, the quotient is — 3b, and the remainder is 


— 3ab? — a®b + 4a. 


As the exponent of d in this remainder is still equal to that of 
6 in the divisor, the division may be continued, by multiplying this 
remainder by 4, in order to render the division of the first term 
possible. This done, the remainder becomes 


— 12ab? — 4a2b + 1623, 


which divided by 46? — 5ab + a’, gives the quotient — 3a, whick 
should be separated from the first by a comma, having no con 
nexion with it. The remainder after this division is 


— 19a?b + 19a’. 


Placing this last remainder under the form 19a?(— 6 + a) 
and suppressing the factor 19a?, as torming no part of the com 
mon divisor, the question is reduced to finding the greatest com. 
mon divisor between 


4b?— 5ab +a? and —b+a. ‘* 


Dividing the first of these polynomials by the second, we obtai! 
an exact quotient, —4b+ a: hence, —b+ a, or a—J, is thé 
greatest common divisor sought. 


288. In the above example, as in all those in which the ex 
ponent of the principal letter is greater by unity in the dividen¢ 
than in the divisor, we can abridge the operation by first multi 
plying every term of the dividend by the square of the co-effi 


1) 


nar. X.] GREATEST COMMON DIVISOR. 305 
cient of the first term of the divisor. We can easily see that by 
‘this means, the first partial quotient obtained will contain the first 
/power of this co-efficient. Multiplying the divisor by the quotient, 
and making the reductions with the dividend thus prepared, the 
result will still contain the co-efficient as a factor, and the division 
can be continued until a remainder is obtained of a lower degree 
than the divisor, with reference to the principal letter. 

| Take the same example as before, viz., 

— 363 + 3ab? — a2b +3 and 452 — 5ab + @?, 

and multiply the dividend by the square of 4 = 16; and we have 


First Operation. 


— 4803 + 48ab? — 16a2b + 1623 | 462 — 5ab + at 
— 12ab2— 40%), + ite — 12b — 3a 
Ist. remainder, — 19a*b + 19a3 ‘ 


or, 19a? (— b+ a). 


(| 


. Second Operation. 


em 80) ba? | b+ta 
— Dial mea 


2d remainder, ae Uh, 
{] 








289. When the exponent of the principal letter in the divi-. 
lend exceeds that of the same letter in the divisor by two, three, 
Ye. units, multiply the dividend by the third, fourth, &c. power 
of the co-efficient of the first term of the divisor. It is easy to 
see the reason of this. 

It may be asked if the suppression of the factors, common to 
ll the terms of one of the remainders, is absolutely necessary, or 
vhether the object is merely to render the operation’ more sim- 
He. It will easily be perceived that the suppression of these fac- 
ors is necessary; for, if the factor 19a? was not Suppressed in 
he preceding example, it would be necessary to multiply the whole 
lividend by this factor, in order to render its first term divisible 
‘y the first term of the divisor; but then, a factor would be in- 
toduced into tne dividend which is also contained in the divisor ; 
nd consequently, the required greatest common divisor would be 


jombined with the factor 19a?, which forms no part of it. 
20 


[ 













306 ELEMENTS OF ALGEBRA. [CHAP. X 


290. For another example, let it be required to find the great- 
est common divisor between the two polynomials, 
at + 3a3b + 4a2b? — 6ab3 + 204 and 4a%b + 2ab? — 26°, 


or simply, 
2a? + ab — 0, 


since the factor 2b can be suppressed, being a factor of the sec- 


ond polynomial and not of the first. 


First Operation. 


Bat + 24a3d + 32a2b2 — 48ab3 + 1654 | 2a? + ab — Bihia 
+ 20a3b + 36a?b? — 48ab3 + a 4a? + 10ab + 130? 
bh 26422 — 38ab3 + 1604 
ast remainder, = Slab? + 2954 
or, — 03 (5la — 296). 
Second Operation. 
Multiply by 2601, the square of 51. 


5202a? + 2601ab — 260182 |) 5la — 295 

5202u2 — 2958ab- 102a +-1096 
lst remainder, 4+ 5559ab — 260182 
5559ab. — 31610? 


2d remainder, + 56082. 


The exponent.of the letter a in the dividend, exceeding that o 
the same letter in the divisor by ¢wo units, the whole dividen 
is multiplied by the cube of 2= 8. ‘This done, we perform thr 
consecutive divisions, and obtain for the first principal remaindes 


— 51ab? + 296+. 


Suppressing 0°, the remainder becomes, — 5la +29); ay 
changing the signs, which is permitted, we have 5la — 296; 
the new dividend is 

2a" + ab — 6?. | 


Multiplying the dividend by the square of 51 = 2601, then ¢ 
fecting the division, we obtain for the second principal remainde 
+ 56052. Now, it results from the second principle (Art. 286 
that the greatest common divisor must be a factor of the remair 
der after each division; therefore it should divide the remaindé 


| 
{ 











RHAP. X.] GREATEST COMMON DIVISOR. 307 


15007. But this remainder is independent of the principal letter a: 
ance, if the two polynomials have a common divisor, it must be 
‘dependent of a, and will consequently be found as a factor in the 
-efficients of the different powers of this letter, in each of the 
|oposed polynomials. But it is evident that the co-eflicients of 
|ese polynomials have not a common factor. Hence, the two 
Nien polynomials are haa with respect to each other. 


Remarks on the greatest common Dwivisor. 


} 

I i The rule for finding the greatest common divisor of two 
: lynomials, may readily be extended to three or more polynomials. 
or, having the polynomials A, B, C, D, &c., if we find the great- 
(st common divisor of A and B, and then the greatest common 
visor of this result and C, the divisor so obtained will evidently 
» the greatest common divisor of A, B, and C; and the same 
‘ocess may be applied to the remaining polynomials. 


|/292. Let A be a rational and entire polynomial, supposed to 
arranged with reference to one of the letters involved in it, a, 
r example. 

If this polynomial is not absolutely prime, that is, if it can be 
‘composed into rational and entire factors, it may be regarded 
| the product of three principal factors, viz., 


| Ist. Of a monomial factor A’, common to all the terms of A. 
jhis factor is composed of the greatest common divisor of all the 


\merical co-efficients, multiplied by the product of the literal fac- 
\fS which are common to all the terms of the polynomial. 
b 2d. Of a polynomial factor A”, independent of a, which is com- 
jm to all the co-efficients of the different powers of a, in the 
‘ranged polynomial. 
| 3d. Of a polynomial factor A”, depending upon a, ard in which 
|2 co-efficients of the different powers of a are prime with each 
rer, so that we shall have 

A — AY x ae x Ae 
Sometimes, one or both of the factors A’, A”, reduce to unity, 


|t the above is the general form of rational and entire polyno- 
lals. Hence, their greatest common divisor may assume the form 


Di TY x D” x DY”, 


508 ELEMENTS OF ALGEBRA. [CHAP. X 


DY denoting the greatest monomial common factor, D” the great- 
est polynomial factor independent of a, and D’” the greatest poly- 
nomial factor depending upon this letter. : 

In ‘order to obtain D’, find the monomial factor A’ common to all 
the terms of A. ‘This factor is in general composed of literal fac- 
tors, which are found by inspecting the terms, and of a numerical 
co-efficient, obtained by finding the greatest common divisor of the 
numerical co-efficients in A. 


In the same way, find the monomial B’ common to all the terms 
of B; then determine the greatest factor D’, common to A‘ and B. 


This factor D’ is set aside, as forming the first part of the re- 
quired common divisor. The factors A’ and B’ are also suppressed 
in the proposed polynomials, and the question 1s reduced to find- 
ing the greatest common divisor of two new polynomials which 
do not contain a common monomial factor. 





EXAMPLES. : 


1. It is required to find the greatest common divisor of the 
two polynomials 


azd2 — ced? — ac? + ct, and 4a%d — 2ac? + 203 — dacd. 


‘he second contains a monomial factor 2. Suppressing it, an¢ 
arranging the polynomials with reference to d, we have 


(a2 — c?) d? — ac? + c4, and (2a? — 2ac)d — ac? + 03. 


It is first necssary to ascertain whether there is a commo 
divisor independent of d. | 

By considering the co-efficients a?.— c? and — atc? + ct, 0 
the first polynomial, it will be seen that — a’c* + c* can be pu 
under the form — c? (a? — c?): hence a2 ¢2 is a common fac 
tor of the co-efficients of the first polynomial, In like manne} 
the co-efficients of the second, 2a? — 2ac and — ac? + c®, ca 
be reduced to 2¢4(a—c) and —c?(a—c); therefore, a—c 
a common factor of these co-efficients. 


Comparing the two factors a? — c? and*a —c, we see that ti 
last will divide the first; hence it follows that a —e is a con 
mon factor of the proposed polynomials, and it is that part o 


their greatest common divisor which is independent of d 
A 


[MHAP. X.] GREATEST COMMON DIVISOR. 309 






} Suppressing a? — c* in the first polynomial, and a—ec in the 
econd, we obtain the two polynomials d? — c? and 2ad — c?, to 
‘vhich the ordinary process must be applied 


d? — ¢? |2ad — 
4a2d? — 4a2c? | 2ad + c? 
+ 2ac?d — 4a2c? 


. 


| — 4a%c? + c+, 
After having multiplied the dividend by 4a, and performed two 
/ onsecutive divisions, we obtain a remainder — 4a2c? + c*; inde- 


‘endent of the letter d: hence the two polynomials d? — c? and 
‘ad — c”, are prime with each other. Therefore, the greatest com- 
ion divisor of the proposed polynomials is a — c. 

| Again, taking the same example, and arranging with reference 
D a, it becomes, after suppressing the factor 2 in the second 
olynomial, 

(d? — c?) a? — cd? 4 ct, and 2da? — (2cd + 0c?) a+ 03. 

| It is easily perceived, that the co-efficient of the different powers 


)f a in the second polynomial, are prime with each other. In the 


|rst polynomial, the co-efficient — c?d? + ct, of the second term, 
|r of a°,-becomes — c? (d? — c?); whence, d? — ¢? is a common 
jictor of the two co- efficients, and since it is not a factor of the 
‘acond polynomial, it may be suppressed in the first, as not form- 
jig a part of the common divisor. 

By suppressing this factor, and taking the second poly neni 
ja dividend and the first for a divisor (in order to avoid prepa- 
jution), we have 


Ist. 2da? — 2cd a+ c3 || @ — 








vie | oun y 
Reme - — 2cd Ja + 2de2 

2 er: 
Or, - 2 a—e, 


7 suppressing the common factor (— 2cd — c?). 

| 2d. a? — ¢? 

~ DESO: —*c? 
bee 


) After having performed the first division, a remainder is ob- 
lined which contains — 2cd — ¢?, as a factor of its two co- 


| oe 


ate 








cd 


310 ELEMENTS OF ALGEBRA. [CHAP. x, 


efficients; for 2dc? + c? = — ¢(— 2d — ¢). This factor being 
suppressed, the remainder 1s reduced to a—e, which will exactly 
divide a? — c?. | 

Hence, a—c is the required greatest common divisor. 


293. There is a remarkable case, in which the greatest com 
mon divisor may be obtained more easily than by the general 
method; it is, when one of the two polynomials contains a letter 
which is not contained in the other. 

In this case, it is evident, that the greatest common divisor ig 
independent of this letter. Hence, by arranging the polynomia 
which contains it, with reference to this letter, the required com 
mon divisor will be the same as that which ewists between the co-effi 
cients of the different powers of the principal letter and the secon 
polynomial. | 

By this method we are led, it is true, to determine the greates 
common divisor between :three or more polynomials. But the 
will be more simple than the proposed polynomials. It often hap 
pens, that some of the co-efficients of the arranyed polynomit 
are monomials, or, that we can discover by simple inspection thé 
they are prime with each other; and, in this case, we are cel 
tain that the proposed polynomials are prime with each other. 

Thus, in the example 1, treated by the first method, after havin 
suppressed the common factor a — ¢, which gives the results, — 

d2—c? and 2ad — e?, : 
we know immediately that these two polynomials are prime wit 
each other; for, since the letter a is contained in the second { 
not in the first, it follows from what has just been said, that t 
common divisor must divide the co-efficients 2d and — c’, whit 
is evidently impossible; hence, they are prime with respect | 
each other. | 


2. Let it be required to find the greatest common divisor of ti 
two polynomials, | 
3bceq + 30mp + 18be + Smpq 

and 4adg — 42fg + 24ad— 7fgq, 


by the last principle. 
We observe. in the first place, that the two polynomials do ' 


contain any common monomial factor. 

























SHAP. X.] GREATEST COMMON DIVISOR. ‘ $1l 


Since g is common to the two polynomials, we can#arrange 
them with reference to this letter, and follow the ordinary rule. 
But as 4 is found in the first polynomial and not in the second. 
If then, we arrange the first with reference to b, which gives 


(3cqg + 18c) b + 30mp + 5mpoa, 


ithe required greatest common divisor will be the same as that which 
jexists between the second polynomial and the two co-efficients 


3cqg-+ 18¢ and 30mp + 5mpq. 


Now, the first of these co-efficients can be put under the form 
3c(q + 6), and the other becomes 5mp(g+ 6); hence g+6 is 
a common factor of these co-efficients. It will therefore be suffi- 
cient to ascertain whether g +6, which is a prime divisor, is a 
factor of the second polynomial. 

Arranging this polynomial with reference to g, it becomes 


F 4 (4ad —- Tf) g — 42fg + 24ad; 


as the second part, 24dd — 42fz = 6 (4ad — 7fz), it follows that 
‘this polynomial is divisible by g+ 6, and gives the quotient 
4ad — 7fg. Therefore, g + 6 is the greatest common divisor of 
the proposed polynomials. 


Remark.—It may be ascertained that g+6 is an exact divisor 
of the polynomial 


(4ad — 7fo)q + 24ad — 42foe, 


by a method derived from the property proved in Art. 278. 
Make ¢+6=0, or g= — 6, in this polynomial; it becomes 


(4ad — 7fg) xX — 6 + 24ad — 42fo — 0; 


that is, —6 substituted for g reduces the polynomial to 0; hence 
g-+ 6 is a divisor of this ‘polynomial. 

This method may be advantageously employed in nearly all the 
applications of the process. It consists in this, viz., after obtain- 
ing a remainder of the first degree with reference to a, when a 
is the principal letter, make this remainder equal to 0, and deduce 
the value of a from this equation. 

If this value, substituted in the remainder of the 2d degree, 
destroys it, then the remainder of the lst degree, simplified Art. 
292, is a common divisor. If the remainder of the 2d degree 


Pod 


312. : ELEMENTS OF ALGEBRA. [CHAP. X. 


does not reduce to 0 by this substitution, we may conclude that 
there is no common divisor depending upon the principal letter. 

Farther, having obtained a remainder of the 2d degree, with 
reference to a, it is not necessary to continue the operation any 
farther. For, 

Decompose this polynomial into two factors of the 1st degree, 
which is-done by placing it equal to 0, and resolving the result- 
ing equation of the 2d degree. 

When each of the values of a thus obtained, substituted in the 
remainder of the 3d degree, destroys it, it is a proof that the re- 
mainder of the 2d degree, simplified, is a common divisor; when 
only one of the values destroys the remainder of the 3d degree, 
the common divisor is the factor of the Ist degree with respect 
to a, which corresponds to this value. 

Finally, when neither of these values destroys the remainder of 
the 3d degree, we may conclude that there is not a common divi- 
sor depending upon the letter a. 

It is here supposed that the two factors of the Ist degree with 
reference to a, are rational, otherwise it would be more simple to 
perform the division of the remainder of the 3d degree by that cf 
the second, and when this last division cannot be performed ex- 


> 


actly, we may be certain that there is no rational common divisor, 
for if there was one, it could. only be of the Ist degree with re- 
spect to a, and should be found in the remainder of the 2d degree, 
which is contrary to the hypothesis. * 


3. Find the greatest common divisor of the two polynomials 
6x° — 4x4 — lla? — 3x? — 3a — 1 
and 4x4 + 223 — 18x27 + 38x — 5. 
Ans. 2x3 — 47? + x — 1 
4. Find the greatest common divisor of the polynomials 
20x® — 12x05 + 162+ — 1523 + 144? — 15a” + 4, 
and 15x4* — 9x3 + 47x? — Qla + 28. 
Ans. 5x7? — 32+ 4. 
5. Find the greatest common divisor of the two polynomials 
5ath? + 2a3b3 + ca? — 3a2b+ + dea 
and a + 5a3d — a3b? + 5a2bd. 


/ 


Ans. a®@#+ab. ) 


é 
CHAP. x.] TRANSFORMATION OF EQUATIONS. 313 


Transformation of Equations. 


| The transformation of an equation consists in changing its form 
without affecting the equality of its members. ‘The object of a 
Tansformation, is to change an equation from a given form, to 
«mother form that is more easily resolved. 


| _ Furst: Transformation. 


To make the Denominators disappear from an Equation. 


294. If we have an equation of the form 
om + Pam 4+ Qerm 24... Tu+U=0, 


and make i +3 
we shall have, after substituting this value for 2, and multiplying 
dwery term by k”, 

y™ + Phy + Qkyn? + RByn—34 eT, mle Oke), 
im equation of which the co-efficients are equal to those of the 
flven equation, multiplied respectively by h°, k1, #2, h3, k4, de. 
This transformation is principally used to make the denominators 
lisappear from an equation, when the co-efficient of the first term is 
nity. 


_As an example, take the equation of the 4th degree, 


/ ee ee CE, eS tog 








b d vi 1 Ona 
If we make c= 4, 
' being a new unknown quantity and & an indeterminate quantity, 
ve have 
| ay ewh f= 0. 





' 


Now, there may be two cases— 

Ist. Where the denominators 4, d, f, h, are prime with each 
ther. In this hypothesis, as & is altogether arbitrary, take 
= bdfh, the product of the denominators, the equation will then 


ecome 


y* + adfh.y> + ch?df*h? . y? + eb3a8f?h3 .y + gbhtd'fih? = 0, 
2 We the co-efficients are entire, and that of its first term 
nity 


Ve 


314 ELEMENTS OF ALGEBRA. (CHAP. X. 


We can determine the values of 2 corresponding to those of y 
from the equation, 


y 
b Wee bah’ 

2d. When the dgnominators contain common factors, we shall 
evidently render the co-efficients entire, by making & equal to the 
smallest multiple of all the denominators. But we can simplify 
still more, by giving to # such a value that A}, #?, h’,. -. shall 
contain the prime factors which compose 4, d, f, h, raised to pow- 
ers at least equal to those which are found in the denominators. 

Thus, the equation 


5 


| 5 13 
i. SS ee 
ax ibaa 


7 
60° goo 
becomes 
5h 5k 7k3 13K 
Ae Bs eS RGR yy eee me | a. . 
6’ ia! Tso" "song ee 


after making # = +. and reducing to entire terms. 


First, if we make & = 9000, which is a multiple of all the 
other denominators, it is clear, that the co-efficients become whok 
numbers. | 

But if we decompose 6, 12, 150, and 9000, into their factors 
we find 


6 = 2. 3, 12 :24.% 3, y150 = Aes ee 9000 = 23 x 3? x 53 
and by simply making 2 
k = genre  X< Oy 
the product of the different simple factors, we obtain . 
k2 = 2? x B2.x 52, B= 23 x 33 x 53) i eee x oe 


whence we see that the values of hk, k?, k3, #4, contain the prim 
factors of 2, 3, 5, raised to powers at least equal to those whic 
enter into 6, 12, 150, and 9000. 

Hence, the hypothesis 


Kc Qi Xd KaOs 


is sufficient to make the denominators disappear. Substituting th: 
value, the equation becomes 


y! 5.2.3.5 ; 5.27.37.57 TBH 13.2%3* 54 


28) 0, 3.7 23.57 “= eee 





P| 
= | 
yy 


 OHAP. X.] TRANSFORMATION OF EQUATIONS. 315 
- 


which reduces to 
yt — 5.5y9 + 5.3.52y? — 7.22.32.5y — 13.2.32.5 — 0; 
or yt — 25y? + 375y? — 1260y — 1170 = 0. 
__ Hence, we perceive the necessity of taking & as small a number 
_as possible: otherwise, we should obtain a transformed equation, 


having its co-efficients very great, as may be seen by reducing 
| the transformed equation resulting from the supposition & = 9000. 


Hence we see, that any equation may be transformed into another 
equation, of which the roots shall be a multiple or sub-multiple of 
those of the given equation. 


EXAMPLES. 


Making « = “. and we have 


| y? — 14y?+ lly — 75 = 0. 
| 13 








21 32 43 l 
| 2. Be Asay ed a ee Ph eee akties ME) ee 
) M40” ns, * G00. S00 
Making x= y = Dds and we have 


| OUEST 
y® — 65y* + 18903 — 307207? — 928800y + 972000 = 0. 


Second Transformation. 
To make the second Term disappear from an Equation. 


295. The difficulty of resolving an equation generally diminishes 
with the number of terms involving the unknown quantity. Thus 
the equation 
‘ 2? =, gives immediately, # = + V 4 
while the complete equation 
x? + 2nr +¢ = 0, 
“requires preparation before it can be resolved. 
__ Now, any given equation can always be transformed into another 
equation,»in which the second term shall be wanting. 
For, let there be the general equation 


| amt Pym-1l 4 Qym24 ..,.4 7% +U=0. 
Suppose a2=u+2’, 


316 ELEMENTS OF ALGEBRA. — [CHAP. X. 
u being unknown, and a’ an | indeterminate quantity. By substitu- 
ting u+ a’ for x, we obtain 

(ut a2’)™+P(u+2’)m> + Q. (ay b ia) hit +T(u+a)+U=0. 


Developing by the binomial formula, and arranging according to 
the decreasing powers of wu. we have 





m—1l 
u™ + max’ jum t+ m. v2 |um 2 4+... tom 


+P + (m—1) Px + Px’ 
+Q + Qu’m2 
ri ries 
4. Ty 
+U 


Since 2’ is entirely arbitrary, we may dispose of it in such a 
way that we shall have 
vs 
mx’ + P=0; whence, o = oe 
Substituting this value of x in the last equation, we shall obtain 
an equation of the form, 


um + Q’um 2 + Rum3 +... Tu+ WW =), 


in which the second term is wanting. 
If this equation were resolved, we could obtain any value of 
corresponding to that of u, from the equation . 


P 
x=ut+ta, or ©T=u——. 
m 


Whence, in order to make the second term of an equation dis- 
appear, 

Substitute for the unknown quantity a new unknown quantity, uni- 
ted with the co-efficient of the second term, taken with a contrary 
sign, and divided by the exponent of the degree of the equation. 


Let us apply the preceding rule to the equation 
x? + 2px = q: 
If we make x= u—p,. 


we have (u — p)? + 2p(u—p)=¢q; 


% 
i 
i 
} 
} 


“CHAP. X.] : Sa OF EQUATIONS. 317 


“and by performing the multiplications and reducing, 





1 uw — p? =4, vt 
which gives u=+/q4+p; 
and consequently, e©=—ptV/q+p?. 


296. Instead of making the second term disappear, it may be 
“required to find an equation which shall be deprived of its third, 
‘fourth, or any other. term. This is done, by making the co-effi- 
, cient of u corresponding to that term equal to 0. For example, 
“to make the third term disappear, we make, in the above-transformed 


_ equation 

lie. m— | 
} 2 
from which we obtain two values for 2’, which substituted in the 
transformed equation reduce it to the form 

i um + Pum + Rum Tut =O. 

Beyond the third term it will be necessary to resolve an equa- 
|tion of a degree superior to the second, to obtain the value of a’ : 
‘and to cause the last term to disappear, it will be necessary to 
“resolve the equation 

afm + Pam... 4+ Te’ + U=0, 
which is what the given equation becomes when 2’ is substituted 
‘Wor a. 
‘eit may happen that the value 








a’? + (m — 1) Px’ + Q=0, 


P 
Y= 


m 


/ 
? 


/which makes the second term disappear, causes also the disap- 
pearance of the third or some other term. For example, in order 
\that the third term may disappear at the same time with the second, 
it is necessary that the value of « which results from the equation 





| ) P 
) c=—-—~) 
m 
shall also satisfy the equation 
) ee fn} 1) Petit, Q2E0. 
Now, if in this last equation, we replace a’ by Jae we have 
jada 20 A ae Q=0, or (m—1) P? —2mQ=0; 


2 





318 ELEMENTS OF sorana lp [CHAP. X. 


and consequently, if 

Pr Ls alin 
m—1 
the disappearance of the second term will also involve that of the 


the third. 


Formation of deriwed Polynomials. 


297. The relation 
e=ut+?2, 


which has been used in the two preceding articles, indicates 
that the roots of the transformed equations are equal to those of 
the given equation, increased or diminished by a certain quan- 
tity. Sometimes this quantity is introduced into the calculus, as 
an indeterminate quantity, the value of which is afterward determined 
by requiring it to satisfy a given condition; sometimes it is a par- 
ticular number, of a given value, which expresses a constant dif- 
ference betweens the roots of a primitive equation and those of 
another equation which we wish to form. 

In short, the transformation, which consists in substituting w + a’ 
for x, in a given equation, is of very frequent use in the theory 
of equations. There is a very simple method of obtaining, in prac- 
tice, the transformation which results from this substitution. 

To show this, let us substitute for 2, u-+ a’ in the equation 


am +. Pym1 4 Qym-2 4+ Rom 34... Tr +U=0; 


then, by developing, and arranging the terms according to the 
ascending powers of u, we have 





a/m + ma/m1 ut nm am -2 u?—+ ym 
+ Px’ +-(m—1) Pa’? +- (m ec —— Py/m-3 
+ Qu’™-2-4.(m—2)Qa’m-3) 4 (m—2) == Qu’m-s 5 | 
Sie Bhi , 
+Tx +T | | 
+U J 


If we observe how the co-efficients of the different powers of x 
are composed, we shall see that the co-efficient of u°, is what the 


‘HAP. X.] FORMATION OF DERIVED POLYNOMIALS. 319 


irst member of the given equation, becomes when «’ is substituted 
n place of x; we shall denote this expression by X’ 

The co-efficient of u! is formed from the preceding term X’, 
xy multiplying each term of X’ by the exponent of «’ in that term, 
md then diminishing this exponent by unity; we shall denote 
his co-efficient by Y’. 

The co-efficient of u? is formed from Y’, by multiplying each 
erm of Y’ by the exponent of a in that term, dividing the prod- 


act by 2, and then diminishing each exponent by unity. Repre- 
: ae \ ; Z’ a 
senting this co-efficient by aay wa see that Z’ is formed from Y’, 


in the same manner that Y’ is formed from X’. 

. In general, the co-efficient of any power of u, in the above- 
wansformed equation, may be found from the preceding co-efficient 
in the following manner: viz., 

| By taking each term of that co-efficient in succession, multiplying 
it by the exponent of x’, dividing by the number which marks the 
olace of the co-efficient, and diminishing the exponent of x’ by unity. 
The law by which the co-efficient 


y 


a ae Teh A OTe 
are derived from each other, is evidently the same as that which 
governs the formation of the terms of the binomial formula (Art. 
203). The expressions, 
ae 6 le Ae ile eee 
~~ 
are called derived polynomials of X’, because each is derived from 
the one which precedes it, by the same law as that by which Y’ 
is deduced from X’. Hence, generally, 
= A derived polynomial is one which is deduced from a given poly- 
nomial, according to a fixed and known law. 
 Recollect that X’ is what the given polynomial becomes when 
w is substituted for «x. 
Y’ is called the first-derived polynomial ; ss 
Z’ is called the second-derived polynomial ; 


| V’ is called the third-derived polynomial. 
&c.. &e. 


320 ELEMENTS OF ALGEBRA. [CHAP. X. 


We should also remember if we make u =0, we shall have, 
aw == 2, whence X’ will become the given polynomial, from which 
the derived polynomials will then be obtained. " 


298. Let us now apply the above principles in the following 


EXAMPLES. 


1. Let it be required to find the derived polynomials from the 
equation 
3at + 623 — 327+ 22 4+1=0=X. 
Now, uw being zero, and a =a, we have from the law of 
forming the derived polynomials, 
X= X= 32t+ 623 — 3227+ 22+1; 
Y’ = 1223 + 182? — Ge +2; 
Z’ = 3627 + 362 —6; 
V’ = 72e@ 4-36; 
YW a 
It should be remarked that the exponent of # in the terms 1, 2 
— 6, 36, and 72, is equal to 0;, hence, each of those terms dis. 
appears in the following derived polynomial. 
2. Let it be required to cause the second term to disappear ir 


the equation 
vt —12x23 + 172? —9r + 7= 0. 


Make (Art. 295), eauto aut 3; 


whence, x” ='3. 
The transformed equation will be.of the form 
7 


xX’ + Yu + Sut + ~—u8 + ut = 0, 


2x | 
and the operation is reduced to finding the values of the co-efficients 
Z’ Vy’ 
XxX’, Y, =~, ==. 
4 were 2.3 


Now, it follows from the preceding law for derived polynomials, 
that 





X’ = (8)*—12 . (39317. (3)?—9. (3) 7, or Xo 10 
Y’ =4.(3)—36 . (3)?+34.(3)'—9, or - =~ Y =—123; 
a = 6 .(3)?—36 . (3)!+17, or - 05 Bey dae = =— 37; 
VW VW’ 

et eC) ee ee | =0. 
Fa aa aie age 






0 

HAP. X.] FORMATION OF DERIVED POLYNOMIALS. 321 

Therefore the transformed equation becomes 
ut — 37u2 — 123u —110=0. 


3. Transform the equation 
4° — 5a? + 7r —9 = 0 
‘uto another equation, the roots of which shall exceed those of 


‘he given equation by unity. 
Make, whence 2 = —1; 


e=u—l; 
7 the transformed equation will be of the form 


7 


/ 
as Yc ss Lichaw gmt a 





} Hence, we have 
aX? = 41(—1)8— 5.(—-1)??+7.(—1)!—9, or 
- -Y* =99; 





| TE oe ee oe 
| aa zy 
eg Ls 1 tg TE f - = my od < x Me 6 Meg 
y= 5 | : 17; 
1 y ’ y’ 

2 ‘ we " - - . ~ WE Van il 4. 


i——-—- 4 - - 
Therefore, the transformed equation becomes 
| 4u — 17u? + 29u — 25 = 0. 


4. What is the transformed equation, if the second term be made 


i disappear in the equation 

| #5 — 10a* + 703 + 42 —-9 = 0? 

. Ans. v5 — 33u3 — 118u? — 152u — 73 = 0. 
5. What is the transformed equation, if the second term be made 


‘o disappear in the equation 
Ba3 + 15x? + 254 —-3 = 0? y 

152 oa 

Sears See Mat 4 


a 
| 6. Transform the equation 
, Sat — 133 + 72? — 82 —9 = 0 


lato another, the roots of which shall be less than the roots of the 





= 


given equation by ia: 
. 102 
Ans. 3u* — 9u3 — 4u? — eu ee Ts ay 


21 


322 ELEMENTS OF ALGEBRA. [CHAP. X 


Properties of derwed Polynomaals. 


299. We will now develop some of the remarkable properties 
of derived polynomials. 


Let X=a™+ Px™ + Qam?,... Tx +U=0 


be a given equation, and a, b, c, d, &c., its m roots. We shall 
then have (Art. 281), 


xm + Pam 4+ Quam 2... = (x -— a) (a — db) (a —c)... (wn — J). 
Making xe=at+u, 

or omitting the accents, and substituting «+ wu for x, and we have 

(2ntu "+ P(e+uy™ +... =(x+u—a)(e+u—d)...; 

or, changing the order of x and uw, in the second member, and re- 

garding « —a, «x —b,...each as a single quantity, 

(x+uy"+ P(x+u)™)....=(u +2x—a) (u-++-x—b) bee (u-+x—l). 


Now, by performing the operations indicated in the two members, 


we shall, by the preceding Article, obtain for the first member, 
{ 


an Yu + uth am 


X being the first member of the proposed equation, and Y, Z . 
the derived polynomials of this member. 
With respect to the second member, it follows from. Art. 295, 


Ist. That the part RL u°, or the last term, is equal to the 
product (a — a)(a — 6)... (# —/) ‘of the factors of the propésed! 
equation. 

2d. The co-efficient of uw is equal to the sum of the produgy 
of these m factors, taken m— 1 and m—1. 


3d. The co-efficient of wu? is equal to the sum of the products 
of these m factors, taken m— 2 and m— 2; and so on. 

Moreover, since the two members of the last equation are iden: 
tical, the co-efficients of the same, powers are equal. Hence, 


X = (w —a)(x—b)(w—c)...(w@—J), 


which was already known. Hence also, Y, or the first-derived 
polynomial, is equal to the sum of the products of the m factors of 
the first tenree in the proposed equation, taken m — 1 and m — 1; 
or equal to the sum of all the quotients that can be obtained by 


4aP. X.] EQUAL ROOTS. . 323 


viding X by each of the m factors of the first degree in the pro- 
sed equation; that ts, 
) Sg 


Sete f= bee pe op RS 





Z Zé 

Also, 7) that is, the second-derived polynomial, divided by 2, 
, equal to the sum of the products of the m factors of the pro- 
sed equation taken m — 2 and m — 2, or equal to the sum of 
‘e quotients that can be obtained by dividing X by each of the 
ctors of the second degree; that is, 

eZ eX © Xx > aria 

| 2 (w—a)(@—6) © (w—a)(w@—c) °°’ (w@ —k) (x — I)’ 

ad so on. 





Of equal Roots. 


"300. An equation is said to contain equal roots, when its first 
‘ember contains equal factors. When this is the case, the de- 
ved polynomial, which is the sum of the products of the m fac- 
rs taken m—1 and m—1, contains a factor in its different 
rts, which is two or more times a factor of the proposed equa- 
m (Art. 299). 

Hence, there must be a common divisor between the first: member 
: the proposed equation, and its first-derived polynomial. 

It remains to ascertain the relation between this common divi- 


rand the equal factors. 


301. Having given an equation, it is required to discover whether 
jhas equal roots, and to determine these roots if possible. 

'Let us make 

A =o" + Px™1 4+ Qor™?+...4+ Tr + 0=0, 

d suppose that the second member contains n factors equal to 
l—a, n’ factors equal to « — 6, n” factors equal to wa—c...; 
d also, the simple factors «—p, x — q, «—r...3; we shall 
mn have, 

(ae — a) (x — 5)” (2. —.c)’... .(@ — p) (2 —¢) (x —17) (1), 
‘We have seen that Y, or the derived polynomial of X, is the 


m of the quotients obtained by dividing X by each of the m fac- 
ts of the first degree in the proposed equation (Art. 299). 










324 ELEMENTS OF ALGEBRA. (CHAP. X, 


Now, since X contains n factors equal to x — a, we shall have 


2% 
n partial quotients equal to ———,; and the same reasoning ap- 


plies to each of the repeated factors, 2 —b, w—c.... More 
over, we can form but one quotient for each simple factor, which 
is of the form, 


xX se x 


e—p 2—-g 2£—Tr 





Therefore, the first-derived polynomial is of the form, 


nX vX n/X X X +g 
ste duo eames 


x—-a 42—b u—e L—p «L—@ 


y= 

















. (2) 


By examining the form of the value of X in equation (1), it is 

plain that 

(a— a)", (2 — b)—1, (a oye 
are factors common to all the terms of the polynomial; hence the 
product 

(@—a)"? X (2 6) Yk (2 ee 
is a common divisor of Y. Moreover, it is evident that it wil 
also divide X: it is therefore a common divisor of X and Y; an¢ 
it is their greatest common divisor. 

For, the prime factors of X are x —a, x —b, w—c..., and 
t—p, ©—q, ©—T...; now, x—p, «—gq, x —7, cannot di 
vide Y, since some one of them will be wanting in each of th 
parts of Y, while it will be a factor of all the other parts. . 

Hence, the greatest common divisor of X and Y is 


D = (a#— a)" (a= b6)”—1 (2 + ¢)/ oe Pe thats, 
The greatest common divisor is composed of the product of thos 


factors which enter two or more times in the given equation, eac! 
raised to a power less by unity than in the primitive equation. 


302. From the above we deduce the following method for find 
ing the equal roots. 
To discover whether an equation 
a= 0 
contains any equal roots, form Y or the derived polynomial of X 
then seck for the greatest common divisor between X and Y; if 


one cannot be obtained, the equation has no equal roots, or equa 
factors. 


i 
| 


HAP. X.] EQUAL ROOTS. 325 
}. 

If we find a common divisor D, and it is of the first degree, 
rof the form x—h, make x —h=0, whence 2 —h. We then 
onelude, that the equation has two roots equal to h, and has but 
ne species of equal roots, {rom which it may be freed by dividing 
¢ by (« — h)?. 

If D is of the second degree with reference to a, resolve the 
quation D=0. There may be two cases; the two roots will 
e equal, or they will be unequal. 

Ist. When we find D = (« — h)?, the equation has three roots 
qual to h, and has but one species of equal roots, from which it 
an be freed by dividing X by (x — h)3. 

2d. When D is of the form (« — h) ( — h’), the proposed 
quation has two roots equal to h, and two equal to h’, from which 
» may be freed by dividing X by (« — h)? (a — W’)?, or by D?. 

_ Suppose now that D is of any degree whatever; it is necessary, 
a order to know the species of equal roots, and the number, of 
pots of each species, to resolve completely the equation 
| Diced: 
| Then, every simple root of D will be twice a root of the given 
gquation ; every double root of D will be three times a root of the 
‘iven equation ; and so on. 

_As to the simple roots of 


. 0. 0, 
ve begin by freeing this equation of the equal factors contained 


1 it, and the resulting equation, X’ = 0, will make known the 
imple roots. 


EXAMPLES. 


| 1. Determine whether the equation 

2x ee 1924. +. Sati Dig 0 
ontains equal roots. 
We have for the first-derived polynomial (Art. 297), 


8x3 — 36x? + 38x — 6. 


] 
Now, seeking for the greatest common divisor of these poly- 
lomials, we find 
F D=x—3=0, whence +=3; 
‘ence, the given equation has two roots equal to 3. 


326 ELEMENTS OF ALGEBRA. [CHAP. X 
Dividing its first member by (« — 3)’, we obtain 
2x2 + 1=0; whence 2=— +2. 
The equation, therefore, is completely resolved, and its roots are 
aAi'3; +o V-2 and ~= V2 


2. For a second example, take 
x —Qet + 323 — 7x? + 8a —3 = 0. 
The first-derived polynomial is ) 
5at — 823 + 9x? — 142 + 8; 
and the common divisor, 
x? —224-+1 = (a —1): 
hence, the proposed equation has three roots equal to 1. 
Dividing its first member by 
(x — 13 = a — 32? + 3x2—1, 
the quotient is 


—l+~7J-—ll 
a +a+3=0; whence t= : 


| 


thus, the equation is completely resolved. 

3. For a third example, take the equation 

ot + 528 + 6x5 — Gat — 15x23 — 32? + Ba + 4 = 0. 
The first-derived polynomial is 
7x8 + 30x25 + 30xt — 2423 — 45x? — 62 + 8; 
and the common divisor is 
at + 323 + a? — 3a — 2. 
The equation 
at + 323 + ao? — 3x —2 = 0 


cannot be resolved directly, but by applying the method of equ: 
roots to it, that is, by seeking for a common divisor between it 
first member and its derived polynomial 

4x3 + 9x? + 22 — 3, 


we find a common divisor, x-+ 1; which proves that the squai 
of «+1 is a factor of at + 323+ x? — 3x2 — 2, and the cu! 
of «+1, a factor of the first member of the given equation. 


i 
| 


CHAP. X.] ELIMINATION. 327 


Dividing 
“at + 303 + wo? —32—2 by (1+ 1)? =a? 4+ 2 + 1, 
‘we have a + x— 2, which being placed equal to zero, gives 
the two roots #—1, «= — 2, or the two factors, «—1 and 
x+2. Hence we have 
at 303+ x? — 32 — 2 = (x + 1)? (wn — 1) (@ + 2). 

Therefore, the first member of the proposed equation is equal to 
| (@ + 1)°(e — 1)? (@ + 2)%3 
that is, the proposed equation has three roots equal to —1, two 
equal to + 1, and two equal to — 2. 

4. What are the equal factors of the equation 


pay a! — Ta? + 10x + 2204 — 4329 — 3527 + 48x + 36 = 0. 

Ans. (x — 2)? (x — 3)? (a+ 1)? = 0. 

ay 5. What are the equal factors in the equation 

x? — 3x6 + 9x5 — 19x* + 2723 — 33a? + 27 —9O = O. 
Ans. (« — 1)3 (a? + 3)? = 0. 


’ Elimination. 


303. To eliminate between two equations of any degree what- 
ever. involving two unknown quantities, is to obtain, by a series of 
operations, performed on these equations, a single equation which 
contains but one of the unknown quantities, and which gives all the 
values of this unknown quantity that will, taken in connexion with 
the corresponding values of the other unknown quantity, satisfy at 
the same time both the given equations. 

This new equation, which is a function of one of the unknown 
quantities, is called the final equation, and the values of the un- 
‘known quantity found from it, are called compatible values. 


Elimination by Means of Indeterminate Multipliers 


| 304. Let there be the equations 
axz+by—c =), 
a’x + b’y —c’ = 0. 
_ If we multiply the first by m, and subtract the second from the 


product, we have 
‘% (ma — a’)x+ (mb —W’/)y—m+e=0...(1). 


328 ELEMENTS OF ALGEBRA. [CHAP. xX. | 


Now, since the value of m is entirely arbitrary, we may give 
it such a value as to render the co-efficient of x zero, which gives 


/ 
a 
ma—a’=0, whence m= mates: 














and (mb — b’)y—m+e¢=0..... (2). 
Substituting in equation (2) the value of m, and we have 
a’ 
—.c—c¢ 
a _@ce—act ac’—a’e 
mer ah —ab’ ab’ —a’b 


Had we chosen to attribute to m such a value as to render the | 
co-efficient of y zero in equation (1), we should have had 


b/ 
mb —b’=0, whence m= 6 
and (ma — a’) x — me+eo = On. Gat), 
Substituting in equation (3) the value of m, we obtain 
b’ 
5 ea 
“ aS , al —a’b 
ae -a—a@ 


The above values for x and y are the same as those deter- 
mined in Art. 97. The principle explained above is applicable 


to three or more equations, involving a like number of unknown 
quantities. 


305. Of all the known methods of elimination, however, the 


method of the common divisor is, in general, the best; it is’ this 
method which we are going to develop. 


Let PACs y) = "Se Ar Bie (x, y) =O) a B, 
be any two equations whatever, in which f and f’ denote any func- | 
tions of x and y. 

Suppose the final equation involving y obtained, and let us try 


to discover some property of the roots of this equation, which may 
serve to determine it. 


Let Vis 
be one of the values of y which will satisfy both the given equations. 
This is called a compatible value of y. It is plain, that, since this 
value of y, in connexion with a certain value of a, will satisfy both 
equations, that if it be substituted in them, there will result two 


* 


HAP. X.] ELIMINATION. 329 


2quations involving « alone, which will admit of at least one com- 
non value of x; and to this common value there will correspond 
2 common divisor involving « (Art. 279). This common divisor 
will be of the first, or of a higher degree with respect to a, ac- 
sording as the particular value of y =a corresponds to one or 
more values of 2. 

. Reciprocally, every value of y which, substituted in the two equa- 
dons, gives a common divisor involving a, is necessarily a compati- 
ble value, because it then evidently satisfies the two equations at 
he same time with the value or values of x found from this com- 
‘mon divisor when put equal to 0. 



















306. We will remark, that, before the substitution, the first mem- 
bers of the equations cannot, in general, have a common divisor 
which is a function of one or both of the unknown quantities. 
For, let us suppose for a moment that the equations 

| Bee Oo tae. 

are of the form 
Ax D=— 0, BxD=—o. 

'D being a function of x and y. ! 

Making separately D = 0, we obtain a single equation involving 
two unknown quantities, which can be satisfied with an infinite 
number of systems of values. Moreover, every system which ren- 
ders D equal to 0, would at the same time cause A’D, B/D to 
vanish, and would consequently satisfy the equations 


A=0. and. B=0. 


Thus, the hypothesis of a common divisor of the two polyno- 
mials A and B, containing x and y, would bring with it as a con- 
sequence that the proposed equations were indeterminate. There- 
fore, if there exists a common divisor, involving « ani ¥, of the 
‘wo polynomials A and B, the proposed equations will be zndeter- 
minate, that is, they may be satisfied by an infinite number of 
systems of values of « and y. Then there would be no data to 
letermine a final equation in y, since the number of values of y 
S infinite. 
If the two polynomials A and B were of the form 
Aged) BS xD; 

D being a function of x only, we might conceive the equation 
D=0 resolved with reference to x, which would give one or 


{ 


{ 
( 


330 ELEMENTS OF ALGEBRA. (CHAP. XxX. 


more values for this unknown. Each of these values substituted 
in the equations 
A’x D=0 and BxD=0), 
would verify them, without regard to the value of y, since D must 
be nothing, in consequence of the substitution of the value of a, 
Therefore, in this case, the proposed equations would admit of 
a finite number of values for x, but of an infinite number of values 
for y, and then there could not exist a final equation in y. 
Hence, when the equations 
A 0) Bae 

are determinate, that is, when they admit only of a limited number 
of systems of values for x and y, their first members cannot have 
for a common divisor a function of these unknown quantities, Un- 
less a particular substitution has been made for one of these 
quantities. | 


307. From this it is easy to deduce a process for obtaining the 
final equation involving y. . 

Since the characteristic property of every compatible value of 
y is, that being substituted in the first members of the two equa’ 
tions, it gives them a common divisor involving x, which they had 
not before, it follows, that if to the two proposed polynomials, ar- 
ranged with reference to x, we apply the process for finding the 
greatest common divisor, we shall generally not find one. But, by 
continuing the operation properly, we shall arrive at a remainder 
independent of 2, but which is a function of y, and which, placed 
equal to 0, will give the required final equation. For, every value 
of y found from this equation, reduces to nothing the last remain: 
der of the operation for finding the common divisor ; it 1s, then, 
such, thai substituted in the preceding remainder, it will render 
this remainder a common divisor of the first members A and B 
Therefore, each of the roots of the equation thus formed, is a = | 
patible value of y. 


308. Admitting that the final equation may be completely 
solved, which would give all the compatible values, it would | 
ward be necessary to obtain the corresponding values of 2, 
it is evident that it would be sufficient for this, to eh th 
different values of y in the remainder preceding the last, put th 
polynomial involving « which results from it, equal to 0, and find 


- CHAP. X.] ELIMINATION. 33] 


from it the values of x; for these polynomials are nothing more 


than the divisors involving 2, which become common to A and B. 


But as the final equation is generally of a degree superior to 


the second, we cannot here explain the methods of finding the 


values of y. Indeed, our design was principally to show that, 


two equations of any degree being given, we can, without supposing 

the resolution of any equation, arrive at another equation, containing 
‘ dae 7 F 

only one of the unknown quantities which enter into the proposed 


‘ equation. 


EXAMPLES. 


1. Having given the equations 
A=x2®+ay+y—1=0, 
a xt "0, 
to find the final equation in y. 


First Operation. 


x? + 3 jeer ey yt — 1 
x3 + yr? + (y? —l)a@ e—y=Q . 


ee gs eck ye hey 
— yx? — yx — y? + y 
R=2 + 2y3 — y = Ist remainder. 





Second Operation. 
aot ye ged [e+ 29° y 
ola plier Sa x — (2y% — 2y) 
—Qy— yet y—1 / 
— (24° — 2y) 2 — 4y° + by* — 2y" 
R’ = 4y® — 6y + 3y? — 1 = 2d remainder. 





Hence, the final equation in y, is 
4y6 — 6y* + 3y? —1=0. 


If it were required to find the final equation in 2, we observe 
that x and y enter in the same manner into the original equations ; 


_hence, « may be changed into y and y into «, without destroying 
the equality of the members. ‘Therefore, 


4x8 — 62 + 3x? —1=0 


is the final equation in z. 


332 ELEMENTS OF ALGEBRA. [CHAP. X. 


2. Find the final equation in y, from the equations | 
A = 0 — 3yz7+ By2— y+ lhea—y¥+y —2y =), 
B=2?—2ye + y—y=0. 
First Operation. | 
x? — Bye? + By? —y+ I a—yt+y? — 2y || —2ry + yr —y | 


a — 2yv?+ (yr? —y)z rigs Q 
— ya® -bi(2y? 1) ey? + ¥? — 2y 
Tee eye ae 
2 —2Q2y= R. 


Second Operation. 
a? —Qey + y?*— y | a — 2y 
ax? — Qxy wre 
y So y = R’. 


, ‘ 
Hence, y2>—y=0 £ 








is the final equation in y. This equation gives 
y=] and tyic:0. 


Placing the preceding remainder equal to zero (Art. 308), and . 
substituting therein the values of 


oj h)vand hayes 0; 


a Ein 


we find for the corresponding values of a, 
+ = 9) and ee 0; 


from which the given equations may be entirely resolved. 


ee Ee ee ee ee ee 


CHAP. XI.] RESOLUTION OF NUMERICAL EQUATIONS. 333 


CHAPTER XI. 


‘RESOLUTION OF NUMERICAL EQUATIONS INVOLVING ONE OR MORE 
UNKNOWN QUANTITIES. 


309. THE principles established in the preceding chapter, are 
applicable to all equations, whether their co-efficients are numeri- 
‘eal or algebraic, and these principles are the elements which are 


_to be employed in the resolution of equations of the higher de- 
“grees. 


It has been already remarked, that analysts have hitherto been 
able to resolve only the general equations of the third and fourth 


degrees. The general formulas which have been obtained for the 


resolution of algebraic equations of the higher degrees, are so 
complicated and inconvenient, even when they can be applied, 
that the problem of the resolution of algebraic equations, of any 
degree whatever, may be regarded as more curious than useful. 


Therefore, analysts have principally directed their researches to 
the resolution of numerical equations, that is, to those which arise 


from the algebraic translation of a problem in which the given 


quantities are particular numbers. Methods have been found, by 
means of which, the roots of a numerical equation of any given de- 
gree, may always be determined. 
| It is proposed to develop these methods in this chapter. 

To render the reasoning general, we will represent the proposed 
equation by 


X = a + Pom! 4+ Qaom2 4+... = 0. 


‘in which P, Q... denote particular numbers which are real, and 
‘either positive or negative. 


334 ELEMENTS OF ALGEBRA. [CHAP. XT 


First Principle. 


310. If we substitute for x a number a, and denote by A what 
X becomes under this supposition; and again substitute a + wu for 
x, and denote the new polynomial by A’: then, u may be taken 
so small, that the difference between A’ and A shall be less than 
any assignable quantity. 

If now, we denote by B, C, D,..... what the co-efficients 

Z V 

? Bs i eis jl 
shall have for the polynomial X, under the supposition that a + wu 
is substituted for a, | 


A+ Bu + Cu? + Dui +... ute 
equal to A+u(B+ Cu + Dw+.. ut) =— A’, 


(Art. 297), become, when we make =a, we 


Now, the quantity is 
u(B+ Cu+ DwW+.. . url) \ a 

is the difference between A’ and A; and it is required to show 
that this difference may be rendered less than any assignable quan- 
tity, by attributing a value sufficiently small to w. 

Let us take the most unfavorable case that can occur, viz., let 
us suppose that every co-efficient is positive, and that each is equal 
to the largest, which we will designate by K. Then, 


Ku(1 + u + uw? +... um) ye ie -'s url) 5 KN 

and in any other case, OO. BG 
Ku(l+u+w+...u™) >u(&+ But ...u™').4- 
But we have, by Art. 61, 


Ku(l ut w+... um) = Ku (7—*); 





and 





when werd. 


_ This being premised, if we wish the difference between A’ and 
A to be less than any number JN, let us make wu such, that 


inte = or <N, which requires that, u= or < Wo) R 


OMAP. ¥T3] RESOLUTION OF NUMERICAL EQUATIONS. 335 


ind any value of wu which will fulfil this last condition, will satisfy 
he inequality 
KuQl+u+twv+...u™)< nN, 

ud consequently, render 

u(B+ Cu+ Duv?+...u™ <Q; 
he which the inequality is greater even than in the expression 
rbove. 
fe 

Second Principle. 

| 311. If two numbers p and q, substituted in succession in the 
piace of X in a numerical equation, give two results affected with 
lcontrary signs, the proposed equation contains a real root, compre- 





hended between these two numbers. 


| ‘Let us suppose that p, when substituted for x in the equation 
X=0, gives +R, 
and that g substituted in the equation 

| X=0, gives — R’. 


Let us now suppose x to vary between the values of p and g 
by so small a quantity, that the difference between any two cor- 
responding consecutive values of X shall be less than any assign- 
able quantity; in which case, we say that X is subject to the 
law of continuity, or that it passes through all the intermediate 
values between R and — R’. 

Now, a quantity which is constantly finite, and subject to the 
law of continuity, cannot change its sign from positive to nega- 
five, or from negative to positive, without passing through zero: 
ence, there is at least one number between p and g which will 
satisfy the equation 


ee 0. 


‘and consequently, one root of the equation lies between these 


numbers. 


312. We have shown in the last article, that if two numbers be 
substituted, in succession, for the unknown quantity in any equation, 
‘and give results affected with contrary signs, that there will be 
‘at least one real root comprehended between them. We are not, 
however, to conclude that there may not be more than one; nor 


a 


336 ELEMENTS OF ALGEBRA. (CHAP. x1, 


that the substitution, in succession, of two numbers which include 
. . . . «WW 

roots of the equation, will necessarily give results affected with 

contrary signs. | 


a - 


Third Principle. 


313. When an uneven number of the real roots of an equation 
are comprehended between two numbers, the results obtained by subs | . 
stituting these numbers in succession for x, will be affected with con- 
trary signs; but of they comprehend an even number of roots, the’ 
results obtained by their substitution will be affected with the same 
sign. 

To make this proposition as clear as possible, denote by a, 3, . 
c,...those roots of the proposed equation, 
xX = 0, 
which are supposed to be comprehended between p and g, | 
by Y, the product of the factors of the first degree, with reference | 
to x, corresponding to the remaining real roots and to the imoging| 


ary roots of the given equation. 
The first member, X, can then be put under the form | 


j 


(1 — a)(x—b)(m—c)...x Y= X. 


Now, substituting p and g in place of x, we shall obtain the 
two results 


(P — 4) (Pi 4) (Dom €) 2.0) Kok 
(7-2) (¢— 2) (¢— ©) 5. eee 
Y’ and Y” representing what Y becomes, when we replace in 
succession, « by p and g. These two quantities Y’ and Y”, are 
affected with the same sign; for, if they were not, by the second 
principle there would be at least one real root comprised between 
p and g, which is contrary to the hypothesis. { 
To determine the signs of the above results more easily, divide 
the first by the second, and we obtain 


(g — a) (q—b)(q—c)... x Y”’ 


which can be written thus, 


eee 











SOTO EWE Tae mad Y’ 
Ti Mpeg tes  aarrih y7° 


— ee eee 


«a 


| : 
CHAP. XI.) LIMITS OF REAL ROOTS. 337 


i; ‘ : 
. Now, since the roots a, b, c,...are comprised between p and 
% we have 


J 
OL COM es 
Pe c 


and aS Cs Re die DO 


| Ata we deduce 


: 


w 3 

7 a, 5 gee’ b, TL C, . . ° 0, 

P P P < 
= 

and —a, g—b, g—ce,.. 

q q qY S 


Hence, since p—a and g—a are affected with contrary signs, 
‘as well as p—6b and g—b, p—e and g—c..., the partial 
‘quotients 
Hs Lars i. 
| eee Lo eae ee a 

Sere TF Ong. HG 


+9 


7 


yy” 
and Y” are affected with the same sign; therefore, the product 


is essentially positive, since Y’ 





are all negative. Moreover, 


p—@ p—b p-—e Y4 

. 2. a aay x . a yr? 

will be negative, when the number of roots, a, b, c..., compre- 
hended between p and gq, is uneven, and positive when the number 





is even. 
Consequently, the two results 


(p —a)(p—b)(p—c)...x Y%, 
a (g—a)(¢g—4)(g—¢).-. x ¥%, 


will have contrary or the same signs, according as the number 
of roots comprised between p and gq is uneven or even. 
; 


_ Limits of Real Roots. 


314. The different methods for resolving numerical equations. 
consist, generally, in substituting particular numbers in the pro- 
posed equation, in order to. discover if these numbers verify it, or 


whether there are roots comprised between them. But by refleet- 
22 


] 


: 


338 ELEMENTS OF ALGEBRA. (CHAP. XI 


ing a little on the composition of the first member of the general 
equation 

am + Pam + Qam2...4+ Ter + U=0, 
we become sensible, that there are certain numbers, above which 
it would be useless to substitute, because all of them above a 
certain limit, would give positive results. 


315. Let it now be required to resolve the following question: 

To determine a number, which substituted in place of x will ren- 
der the first term x™ greater than the arithmetical sum of all the 
other terms. 

Suppose all the terms of the equation to be ‘negative, except 
the first, so that 

am — Pym — Qr™?,...— Tr —-U=0. 
It is required to find a number for « which will render 


a™ > Px™ + Qam-2 + ...4+ Tx + U. 
Let & denote the greatest co-efficient, and substitute it in place 


of the co-efficients; the inequality will then become 
2” > kam) 4+ kom 2. + he + ik. 

It is evident that every number substituted for « which will 
satisfy this condition, will for a stronger reason, satisfy the pre- 
ceding. Now, dividing this inequality by 2”, it becomes 
k k 


gm—l am 





k k k 
oe ee 
x 2 «3 


Making x =k, the second member becomes J a1 plus a_ 


series of positive fractions. The number & will therefore not sat- 
isfy the inequality; but by supposing «= k-+ 1, we obtain for 
the second member the series of fractions 


ER cl tt ie k k 
PI LIP LEI Ge 


which, considered in an inverse order, is an increasing geometri- 





cal progression, the first term of which is the me 


nae. 
(A+ 1)” 





; hence, the expression for te | 


k +1, and the last term . 
+1 
sum of all the terms is (Art. 192), 


|\HAP. XI.] ORDINARY LIMIT OF POSITIVE ROOTS. 339 


k k 
Se 1 alias SEES 
CE ae 
k+1—1 rm (A+ 1)” 


(rhich is evidently less than unity. 
Now, any number > (k + 1), put in place of x, will ae the 












um of the fractions - ee : tai . still less. Therefore, 


"The greatest saftfiaien’ Ns unity, or any greater number, being 
ubstituted for x, will render the first term x™ greater than the arith- 
ier sum of all the other terms. 


316. Every number which exceeds the greatest of the posi- 
He roots of an equation, ts called a superior limit of the positive 
nets. 

_From this definition, it follows, that this limit is susceptible of 
/n infinite number of values. For, when a number is found to ex- 


4 also a superior limit. 

| But since the largest of the positive roots will, when substituted 
x a, merely reduce the first member to zero, it follows, that 
‘re shall be sure of obtaining a superior limit of the positive roots 
ly finding a number, which substituted in place of x renders the first 
vember positive, and which at the same time is such, that every 
|reater number will also give a positive result. 

Hence, the greatest co-efficient of x plus unity, is a superior limit 
F the positive roots. 


Ordinary Limit of the Positive Roots. 


' 317. The limit of the positive roots obtained in the last article, 
| commonly much too great, because, in general, the equation 
|ontains several positive terms. We will, therefore, seek for a 
mit suitable for all equations. 

| Let a" denote that power of a, corresponding to the first neg- 
Liive term which follows 2”, and let us consider the most unfavor- 
|ble case, viz., that in which all the succeeding terms are negative 
nd affected with the greatest of the negative co-efficients in the 
quation. 

Let S denote this co-efficient. What conditions will render 


mat th Sar ule Sat -B 87 





340 ELEMENTS OF ALGEBRA. ‘CHAP. XI. 


Dividing both members of this inequality by 2, we have 
S S S S S 
LS cat aa yee om 


ml 


Now, by supposing 
x= %/ S +1, or for simplicity, making 4f S = 








which gives, S aea8’*\ te and. oe =e eas 
the second member of the inequality will become, : 
S/ S/n S/n S/n 
(iy ee TT pe ee 
which is a progression by quotients, of which eae is the 
: 
first term, S’+ 1 the ratio, and ("fay the last term. Hence, 
the expression for the sum of all the terms is (Art. 192), 
Sih ' Sn 
(0S a)e . (S i 1) — (S’+ (S++ 1)" S/n- S/a-l 
TENET TE (8 iy (SF 


which is evidently less than 1. 


Moreover, every number > S’+1 or 4/S +1, will, when 
substituted for xz, render the sum of the fractions | 


S S 
aber 


yn grtl 


| 
still smaller, since the numerators remaining the same, fae de 
nominators will increase. 


Hence, "/ S +1, and any greater number, will render the first 
term a” ereater than the arithmetical sum of all the negative terms. 
of the equation, and will consequently give a positive result for 
the first member. Therefore, : 

Unity increased by that root of the greatest negative co-efficient | 
whose index is the number of terms which precede the first negative| 
term, is a superior limit of the positive roots of the equation. If sh 
co-efficient of a term is 0, the term must still be counted. 

Make n= 1, in which case the first. negative term is the sec- 
ond term of the equation; the limit becomes : 


v Sia Sega 


that is, the greatest negative co-efficient plus unity. 


a 






| SHAP. =r, SMALLEST LIMIT IN ENTIRE NUMBERS. 341 


| Let n =2; then, the limit is VS +1. When n=3, the 
Mimit is VS +1. 


EXAMPLES. 


1. What is the superior limit of the positive roots in the equation 
a* — 5x3 + 37a? — 3x + 39 = 0. 


Anscs? Site) = 4/0: eee 


2 What is the superior limit of the positive roots in the equation 
x + Tat — 1223 — 492? + 5227 —13. = 0. 


3. What is the superior limit of the positive roots in the equation 
a* + lla? — 25¢7 — 67 = 0. 
In this example, we see that the second term is wanting, that is, 
its co-efficient is zero; but the term must still be counted in fix- 
ing the value of n. We also see, that the largest negative co-efi- 
cient of x is found in the last term where the exponent of x is 
zero. Hence, 
PISA ee er tag. 

and therefore, 6 is the least whole number that will certainly ful- 
fil the conditions. 


Smallest Limit in Entire Numbers. 


318. In Art. 316, it was shown that the greatest co-efficient of 
% plus unity, is a superior limit of the positive roots. In the last 
article we found a limit still less; and we now propose to find 
the smallest limit in whole numbers. 


Let . x was 0, 
be the proposed equation. If in this equation we make #x=2+4u, 
x’ being indeterminate, we shall obtain (Art. 297), 


X’ + Yut Sey... fu =0 (1). 


Let us suppose, that after successive trials we have determined 
a number for 2, which substituted in 


: ZL 


Pepe 8S iit ae 


¥ 


342 ELEMENTS OF ALGEBRA. [CHAP. XI. 


renders all these co-efficients positive at the same time; this num-_ 
ber will be greater than the greatest positive root of the equation 


Pi a tS 


For, if the co-efficients of equation (1) are all positive, no posi- 
tive number can verify it; therefore, all of the real values of uw 
must be negative. But from the equation 


e=—a+u, we hae u=2— 2’; 
and in order that every value of u, corresponding to each of the 


values of w and 2’, may be negative, it is necessary that the great- 
est positive value of a should be less than the value of a’. 


EXAMPLES 


Let et — 523 — 62? — 192 +7 = 0. 


As a’ is indeterminate, we may, to avoid the inconvenience of 
writing the primes, retain the letter x in the formation of the de- 
rived polynomials; and we have } 


X = wt— 5e3— 622+ 192 +7, 
Y = 403 — 15x? —12% — 19, 

Z 

or = 642 — 15x —6, 

vs —4 5 

Ne aah: MEA 


The question is now reduced to finding the smallest entire num- 
ber which, substituted in place of a, will render all of these poly- | 
nomials positive. . 

It is plain that 2 and every number >2, will render the poly- 
nomial of the first degree positive. | 

But 2, substituted in the polynomial of the second degree, gives 
a negative result; and 3, or any number > 3, gives a positive 
result. | 

Now 3 and 4, substituted in succession in the polynomial of 
the third degree, give negative results; but 5, and any greater 
number, gives a positive result. 

Lastly, 5 substituted in _X, gives a negative result, and so does 
6; for the first three terms, at — 5x3 — 6x2, are equivalent to the 
expression x*(* — 5) — 6x’, which reduces to 0 when x = 6; but 
«= 7 evidently gives a positive result. Hence 7, which here. 







, ‘ 
, CHAP. XI.} SUPERIOR LIMIT OF NEGATIVE ROOTS. 343 


L.stands for a’, is a superior limit of the positive roots of the given 
equation. Since it has been shown that 6 gives a negative re- 
sult, it follows that there is at least one real root between 6 and 7. 


2. Applying this method to the equation 
2° — 3at — 8x3 — 25a? + 4a — 39 = 0, 

the superior limit is found to be 6. 
_? 3. We find 7 to be the superior limit of the positive roots of 
‘the equation 
x5 — 5a* — 1323 + 1722 — 69 = O. 
_ This method is seldom used, except in finding incommensurable 
‘roots. 





| 
Superior Limit of negative Roots.—Inferior Limit of posi- 


j tive and negatwe Loots. 


319. Having found the superior limit of the positive roots, it 
‘only remains to find the inferior limit, and the superior and infe- 
‘rior limits of the negative roots. 


, 


Let, 2 = superior limit of positive roots. 
I’ = inferior limit of positive roots. 
L/ = superior limit (that is, numerically) of negative roots. 
L/” = inferior limit of negative roots. 
2 1 
_ Ist. If in any equation X = 0, we make » = —, we have a 
| : y 
derived equation Y=0. We know from the relation »=—, that 


the greatest positive value of y will correspond to the smallest 
of x; hence, designating the superior limit of the positive roots 


] 
of the equation Y =O by L, we shall have as L’, the inferior 


limit of the positive roots of the given equation. 

2d. If in the equation X=0, we make x = — y, which gives 
the transformed equation Y= 0, it is clear that the positive roots 
of this new equation, taken with the sign —, will give the nega- 
tive roots of the given equation; therefore, determining, by the 
known methods, the superior limit L of the positive roots of the 
equation Y= 0, we shall have — L = L’/’, the superior limit (nu- 
. merically) of the negative roots of the proposed equation. 


344 ELEMENTS OF ALGEBRA. s (CHAP. XI, | 
3d. Finally, if we replace x, in the given equation, by ap | 
y 
and find the superior limit Z of the transformed equation Y = 0 
1 ; a. ake } 
then, 0°” =" — a will be the inferior limit (numerically) of the 


negative roots of the given equation. 
Consequences deduced from the preceding Principles. 
First. 


320. Every equation in which there are no variations in the SUENS, 


that is, in which all the terms are positive, must have all of its real | 


roots negative; for, every positive number substituted for 2, will 
render the first member essentially positive. 


Second. 


321. Every complete equation, having its terms alternately posi- 
tive and negative, must have its real roots all positive; for, every 


negative number substituted for « in the proposed equation, would | 


render all the terms positive, if the equation was of an even de- | 


gree, and all of them negative, if it were of an odd degree. Hence, 
their sum could not be equal to zero in either case. 


This principle is also true for every incomplete equation, in which | 


there results, by substituting —y for x, an equation having all of 
its terms affected with the same sign. 


Third. ' 


322. Every equation of an odd degree, the co-efficients of which 
are real, has at least one real root affected with a sign contrary to 
that of its last term. 

For, let 

am + Pre +...Tr + U=0, 


be the proposed equation; and first consider the case in which 
the last term is negative. 

By making x = 0, the first member becomes — U. But by 
giving a value to x equal to the greatest co-efficient plus unity, 
or (K + 1), the first term «” will become greater than the arith- 
metical sum of all the others (Art. 315), the result of this sub- 
stitution will therefore be positive; hence, there is at least one 


i 


i 
\ 


th 
(CHAP. XI.J CONSEQUENCES OF PRECEDING PRINCIPLES. 345 


real root comprehended between 0 and K + 1, which root is posi- 
‘ive, and consequently alfected with a sign contrary to that of the 
last term. 

; Suppose now, that the last term is positive. 

Making «—0, we obtain + U for the result; but by putting 
‘— (K + 1) in place of «x, we shall obtain a negative result, since 
che first term becomes negative by this substitution; hence, the 
‘squation has at least one real root comprehended between 0 and 
ea (K + 1), which is negative, or affected with u-sign contrary to 
chat of the last term. 








I Fourth. 

f 323. Every equation of an even degree, involving only real co- 
efficients of which the last term is negative, has at least two real 
roots, one positive and the other negative. 

| For, let’ — U be the last term; making x = 0, there results 
— U. Now substitute either K+1, or — (K+ 1), K being the 
greatest co-efficient in the equation. As m is an even number, the 
‘frst term a” will remain positive ; besides, by these substitutions, 
it becomes greater than the sum of all the others; therefore, the 
results obtained by these substitutions are both posztive, or affected 
with a sign contrary to that given by the hypothesis x= 0; hence, 
the ‘equation has at least two real roots, one positive, and compre- 
fhended between 0 and K +1, the other negative, and compre- 
hended between 0 and — (K + 1). 


Fifth. 


324. If an equation, involving only real co-efficients, contains !- 


aginary roots, the number of such roots must be even. 
For, conceive that the first member has been divided by all the 
simple factors corresponding to the real roots; the co-efficients 
of the quotient will be real (Art. 278); and the equation must also 
be of an even degree; for, if it was uneven, by placing it equal 
0 zero, we should obtain an equation that would contain at least 


‘me real root; hence, the imaginary roots must enter by pairs. 
Remark.—325. There is a property of the above polynomial 
quotient which belongs exclusively to equations containing only 
imaginary roots; viz., every such equation always remains positive 
for any real value substituted for x. 





; 
. * 


346 ELEMENTS OF ALGEBRA. [CHAP. XI, 


For, by substituting for a, K+ 1, the greatest co-efficient plus 
unity, we could always obtain a positive result; hence, if the 
polynomial could become negative, it would follow that when 
placed equal to zero, there would be at least one real root com- 
prehended between K+ 1 and the number which would give a 
negative result (Art. 311). 


It also follows, that the last term of this polynomial must be 
positive, otherwise « = 0 would give a negative result. 


Sixth. 


326. When the last term of an equation is positive, the number 
of its real positive roots is even; and when tt is negative, the num- 
ber of such roots 1s uneven. 


For, first suppose that the last term is + U, or positive. Since 
by making «—0, there will result + U, and by making x= K + 1, 
the result will also be positive, it follows that 0 and K+ 1 give 
two results affected with the same sign, and consequently (Art, 
313), the number of real roots, if any, comprehended between them, 
is even. 

When the last term is — U, then 0 and K +1, give two re- 
sults affected with contrary signs, and consequently comprehend 
either a single root, or an odd number of them. 


\ 
The reciprocal of this proposition is evident. 


Descartes’ Rule. 


327. An equation of any degree whatever, cannot have a greater 
number of positive roots than there are variations in the signs of its 
terms, nor a greater number of negative roots than there are per 
manences of these signs. : 


A variation is a change of sign in passing along the terms, and) 
a permanence is when two consecutive terms have the same sign, 


In the equation « —a=O, there is one variation, and one posi- 
tive root, «=a. And in the equation « +5=0, there is one. 
permanence, and one negative root, r= — 6. : 

If these equations be multiplied together, there will result an 
equation of the second degree, 
eg } = 0. 


a*—a 


+6 

















» CHAP. XI.) DESCARTES’ RULE. 347 


If a is less than 0}, the equation will be of the first form (Art. 


144); and if a>+b the equation will be of the second form; that is, 


a<b gives 2?+ 2px —q=0, 


and ee rt = Qor eg = 0. 


In the first case, there is one permanence, and one variation, and 
in the second. one variation and one permanence. Since in either 


form, one root is positive and one negative, it follows that there 
‘are aS many positive roots as there are variations, and as many 


negative roots as there are permanences. 
The proposition will evidently be demonstrated in a general 


manner, if it be shown that the multiplication of the first member 
| by a factor # —a, corresponding to a positive root, introduces at 
least one variation, and that the multiplication by a factor x + a, 


corresponding to a negative root, introduces at least one permanence. 
Take the equation 
om + Agml + Bom? + Co™ 35 +... 4 Tr t U=G, 
in which the signs succeed each other in any manner whatever. 
By multiplying by x —u, we have 
eho Ca gate a ke, ects Of 


= Ba 


am + B 


= Aa 


amti+ A x 


croat 1Cb 














+ Ua ae 
The co-efficients which form the first horizontal line of this 
product, are those of the given equation, taken with the same 
siyn; and the co-efficients of the second line are formed from 
those of the first, by multiplying by a, changing the signs, and 
advancing each one place to the right. 
Now, so long as each co-efficient of the upper line is greater 


sree WY 


than the corresponding one in the lower, it will determine the 


sign of the total co-efficient; hence, in this case there will he, 
from the first term to that preceding the last, inclusively, the same 


_ variations and the same permanences as in the proposed equation ; 


but the last term = Ua having a sign contrary to that which im- 
mediately precedes it, there must be one more variation than in 
the proposed equation. 

When a co-efficient in the lower line is affected with a sign 
contrary to the one corresponding to it in the upper, and is also 
greater than this last, there is a change from a permanence of sign 
to a variation; for the sign of the term in which this happens, 


oan * ELEMENTS OF ALGEBRA [CHAP. XI, 


being the same as that of the inferior co-efficient, must be con- 
trary to that of the preceding term, which has been supposed to 
be the same as that of its superior co-efficient. Hence, each 
time we descend from the upper to the lower line, in order to 
determine the sign, there is a variation which is not found in the 
proposed equation; and if, after passing into the lower line, we 
continue in it throughout, we shall find for the remaining terms 
the same variations and the same permanences as in the given 
equation, since the co-efficients of this line are all affected with 
signs contrary to those of the primitive co-efficients. This sup- 
position would therefore give us one variation for each positive 
root. But if we ascend from the lower to the upper line, there — 
may be either a variation or a permanence. But even by sup- — 
posing that this passage produces permanences in all cases, since 
the last term == Ua forms a part of the lower line, it will be ne- 
cessary to go once more from the upper line to the lower, than 
from the lower to the upper. Hence, the new equation must have 
at least one more variation than the proposed; and it will be the 
same for each, positive root introduced into it. 

It may be demonstrated, in an analogous manner, that the mui- 
tiplication by a factor «+ a, corresponding to a negative root, 
would introduce one permanence more. Hence, in any equation, the 
number of positive roots cannot be greater than the number of «— 
VARIATIONS of signs, nor the number of negative roots greater 
than the number of PERMANENCEs. | 


. 


Consequence. 





328. When the roots of an equation are all real, the number of 
positive roots is equal to the number of variations, and the number : 
of negative roots to the number of permanences. | 

For, let m denote the degree of the equation, n the number of 
variations of the signs, p the number of permanences; we shall 
have m=n-+ p. Moreover, let n’ denote the number of positive | 
roots, and p’ the number of negative roots, we shall have 


mi =a -- pes : 
whence nt+p=n +p’, or, n—n =p —p. 
Now, we have just seen that n’ cannot be > 2, nor can it be 


less, since p’ cannot be >>; therefore we must have n’ =n, 
and p’ =p. 


CHAP. XI.] COMMENSURABLE ROOTS OF EQUATIONS. 349 


Remark.—329. When an equation wants some of its terms, we 
can often discover the presence of imaginary roots, by means of 
the above rule. 

For example, take the equation 


| 3+ pxe+gq=0, 
‘? and g being essentially positive ; introducing the term which is 
wanting, by affecting it with the co-efficient +0: it becomes 
| o+0.2?+ prt g=0. 
_ By considering only the superior sign, we should only obtain 
permanences, whereas the inferior sign gives two variations. This 
‘Proves that the equation has some imaginary roots; for, if they 
| were all three real; it would be necessary by virtue of the supe- 
Mior sign, that they should be all negative, and, by virtue of the 
‘inferior sign, that two of them should be positive and one nega- 
‘tive, which are contradictory results. 

We can conclude nothing from an equation of the form 
| 3—pe+g=0; 
‘for, introducing the term + 0.22, it becomes 
e+ 0.2? —pxe+g=0, 
| which contains one permanence and two variations, whether we 
| take the superior or inferior sign. ‘Therefore, this equation may 
have its three roots real, viz., two positive and one negative; or; 










two of its roots may be imaginary and one negative, since its last 
term is positive (Art. 326). 


Of the. commensurable Roots of Numerical Equations. 


330. Every equation in which the co-efficients are whole num- 
‘bers, that of the first term being unity, will have whole numbers 
only for its commensurable roots. 

For, let there be the equation , 

am + Poml4 Qyr24+...4+7Tr+U=0; 
in which P,Q... 7, U, are whole numbers, and suppose that it 


a 
‘were possible for one root to be a commensurable fraction Fz 


ie this fraction for x, the equation becomes 


a 
mt Po t+ OTe t.. + TE+U=0; 


350 ELEMENTS OF ALGEBRA. [CHA?, XT, 


whence, multiplying both members by b”~1, and transposing, 
m= — Pav — Que — ... — Tabm4 — Yor, 

But the second member of this equation is composed of a series 
of entire numbers, while the first is essentially fractional, for a and 
} being prime with each other, a” and b will also be prime with 
each other (Art. 118), and hence this equality cannot exist; for, an 
irreducible fraction cannot be equal to a whole number. 

Therefore, it is impossible for any commensurable fraction to 
satisfy the equation. Now, it has been shown (Art. 294), that an 
equation containing rational, but fractional co-efficients, can be 
transformed into another in which the co-efficients are whole num- 
bers, that of the first term being unity. Hence the research of the 
commensurable roots, either entire or fractional, can always be re- 
duced to that of the entire roots. 


331. This being the case, take the general equation 
a” + Pam + Qam 2... 4+ Re + Se?+ Te + U=0, 
and let @ denote any entire number, positive or negative, which 
will verify it. 
Since a@ is a root, we shall have the equation 
a fe Pama oy Ra + Sa? Ta ee), 
Replace a by all the entire positive and negative numbers be- 
tween | and the limit + ZL, and between —1 and — L”: those 
which verify the above equality will be roots of the equation. But 
these trials being long and troublesome, we will deduce from equa- 
tion (1), other conditions equivalent to this, and easier verified. 
Transposing all the terms except the last, and dividing by a, 
equation (1) becomes 


De get py Li, ee 


a 
Now, the second member of this equation is an entire number; 
iy : 
hence — must be an entire number; therefore, the entire roots of 
i 
the equation are comprised among the divisors of the last term. 
Transposing — T in equation (2), dividing by a, and making 


a +T=T’, we have, 


= — ar? — Par 3... — Ra—S... (3). 











‘HAP. XI.] § COMMENSURABLE ROOTS OF EQUATIONS. 35] 


' ; ™ ae od 3. 
The second member of this equation being entire, —, that is, 
a 


he quotient of the division of 
% + T by a, 






'y an entire number. 
 Transposing the term — S and dividing by a, we have, by 
supposing 


a Sts 
7 
Re. Pym mk on, — Bi sin), 


f 


‘ . : : S : 
The second member of this equation being entire, —, that is, 
a 

the quotient of the division of 
db 
—-+S by a 
a 

‘s an entire number. 


By continuing to transpose the terms of the second member into 
the first, we shall, after m — 1 transformations, obtain an equa- 


tion of the form 


Ke 
ae — —a—P. 
a 
Then, transposing the term — P, dividing by a, and making 
/ / 7 
—+ P= P’, we have Sigh ——/1,, OF titel 1=0 
a . a a 


This equation, which is only a transformation of equation (1), 
is the last condition which it is requisite for the entire number a 
to satisfy, in order that it may be known to be a root of the 
equation. 

332. From the preceding conditions we conclude that, when an 
entire number a, positive or negative, is a root of the given equa- 
tion, the quotient of the last term, divided by a, ts an entire number. 

Adding to this quotient the co-efficient of #', the quotient of this 
sum, divided by a, must also be entire. 

Adding the co-efficient of x to this last quotient, and again divi- 
ding by a, the new quotient must also be entire; and so on. 


352) ELEMENTS OF ALGEBRA. [CHAP. X¥. 


Finally, adding the co-efficient of the second term, that is, of | 
a1, to the preceding quotient, the quotient of this sum divided by © 
a, must be equal to —1; hence, the result of the addition of unity, 
which is the co-efficient of x™, to the preceding quotient, must be’ 
equal to 0. ' a | 

Every number which will satisfy these conditions will be a root,” 
and those which do not satisfy them should be rejected. 

All the entire roo‘; may be determined at the same time, as 
follows: r 

After having determined all the divisors of the last term, write 
those which are comprehended between the limits +L and — L’ upon 
the same horizontal line; then underneath these divisors write the 
quotients of the last term by each of them. 


Add the co-efficient of x! to each of these quotients, and write the 
sums underneath the quotients which correspond to them. Then divide 
these sums by each of the divisors, and write the quotients underneath 
the corresponding sums, taking care to reject the fractional quotients 
and the divisors which produce them; and so on. 


When there are terms wanting in the proposed equation, their 
co-eflicients, which are to be regarded as equal to 0, must be 
taken into consideration. 


Pa EXAMPLES. 


1. What are the entire roots of the equation 
at — x3 — 1302+ 16x — 48 = 0? 


The superior limit of the positive reots of this equation (Art 
317), is 13+1=14. The co-efficient 48 need not be considered, | 
since the last two terms can be put under the form 16 («— 3);_ 
hence, when x > 3, this part is essentially positive. | 

The superior limit of the negative roots (Art. 319), is 


(f+ 4/48). or Ja 

Therefore, the divisors of the last term which may give roots,.| 
are 1, 2, 3, 4, 6, 8, 12; moreover, neither + 1, nor —1, will 
satisfy the equation, because the co-efficient — 48 is itself greater 
than the sum of all the others: we should therefore try only the - 
positive divisors from 2 to 12, and the negative divisors from — 2 
to — 6 inclusively. 









| GHAP. XI.J COMMENSURABLE ROOTS OF EQUATIONS. 353 


By observing the rule given above, we have 


Lyle) a SI 9 OF > top he agg it 
= 4, — 6, —8, —12, —16, —24, +24, +16, +12, + 8 
12, 8 10, ‘peer ef 4, 0, a 8, = 40, “oe, {7 28, + 24 


1, See AE i * 0, Migs ERG 01 Dy ORs ira 4 
ee 12, peer et, — 13, —17, — 33. bh oe a 
= 7 ee oy “i i iM ee a 

_ ae e vn 2D ¥ fy AN 

i can nan »,— l 


gels Ape! SE Rac alin tan ee ARE 


The first line contains the divisors, the second contains the 
‘quotients arising from the division of the last term — 48, by each 
‘of the divisors, The third line contains these quotients augmented 
“by the co-efficient + 16, and the fourth the quotients of these 
‘sums by each of the divisors; this second condition excludes the 
divisors +8, +6, and — 3. 

_ The fifth is the preceding line of quotients, augmented by the 
(co-efficient — 13, and the seth is the quotients of these sums by 
‘each of the divisors; this third condition excludes the divisors 
3, 2, — 2, and — 6. 

Finally, the seventh is the third line of quotients, augmented hy 
the co-efficient — 1, and the eighth is the quotients of these sums 
by each of the divisors. The divisors 14 and — 4 are the only 
ones which give —1; hence, +4 and — 4 are the only entire 
roots of the equation. 


In fact, if we divide 
xt — x3 — 13x? + 162 — 48, 


(by the product (x — 4) (w+ 4), or 2? — 16, the quotient will be 
a? — x -+ 3, which placed equal to zero, gives 
ee 
hae, 2 
therefore, the four roots are 
i 1 
wae 4, e+ 11 and soe V= ih 


_ 2. What are the entire roots of the equation 


xt — 543 + 254 —21=0? 
23 


eT). 


354 ELEMENTS OF ALGEBRA. - (CHAP. XT. 


3. What are the entire roots of the equation 
1525 — 192+ + 623 + 15x? —192 +6=—0? 


yw 4. What are the entire roots of the equation 


S ga6 + 3005 + 2224 + 1023 + 172? — 202 + 4=0. 


ri 
j 


‘ Sturms’ Theorem. 


i 


333. The object of this theorem is to explain a method of de- 
termining the number and places of the real roots of equations in- 
volving but one unknown quantity. Let 


A = OU 


represent an equation containing the single unknown quantity 2; 
_& being a polynomial of the m” degree with respect to a, the 
‘co-efficients of which are all real. If this equation should have 
equal roots, they may be found and divided out as in Art. 302, 
and the following reasoning be applied to the equation which would 
result. We will therefore suppose X — 0 to have no equal roots. 


334. Let us denote the first-derived polynomial of X by X,, and 
then apply to X and X, a process similar to that for finding their 
greatest common divisor, differing only in this respect, that instead 
of using the successive remainders as at first obtained, we change 
their signs, and take care also, in preparing for the division, neither 
to introduce nor reject any factor except a positive one. 

If we denote the several remainders, in order, after their signs 
have been changed, by X,, X,...X;,, which are read X second, 
X third, &c., and denote the corresponding quotients by Q,, Q, 

. Q,-,, we may then form the equations 


X= 2%,0'— xo Ae 
2,5 cao 


iene = Me Q, X nity) pains Cae 
Xp—-g = X,4Q,-1 — X, 
Since by hypothesis, X = 0 has no equal roots, no common di- 


visor can exist between X and X, (Art. 300). The last remainde: 
— X,, will therefore be different from zero, and independent of x. 


AAP. XI.J STURMS’ THEOREM. 355 


'335. Now, let us suppose that a number p has been substituted 
‘t @ in each of the expressions X, X,, X,... X,_,; and that 
e signs of the results, together with the sign of X,, are arranged 
a line one after the other: also that another number q, greater 
an p, has been substituted for #, and the signs of the results 
‘ranged in like manner. | 
, Then will the number of variations in the signs of the first ar- 
‘ngement, diminished by the number of variations in those of the 
‘cond, denote the exact number of real roots comprised between p 


id q. 


"336. The demonstration of this truth mainly sa he upon the 
er following properties of the expressions X, X,...X,, &c. 


‘I. Let @ be a root of the equation X —0. If we substitute 
ae u for x, and designate by A what X becomes, and denote the 
“rived polynomials by A’, A”, A”, &c.; we shall have (Art. 299), 


A” 





A+ A’u + 
But since by hypothesis, a is a root of the equation X — 0, we 
we A =0, and hence the above expression becomes 

: . AY” 


u (A’ + u aries 
| 
i 











Sokgiod ae wed 


which A’ is not zero, since the equation X = 0 is supposed 
‘t to contain equal roots. Now we say, that u can be made so 
iall, that the sign of the quantity within the parenthesis shall be 
‘ same as that of its first term. 

‘We attain this object, by finding for w a value which shall ren- 
* numerically, 








3 4/ 44/ 
A’> a u+ 2 u? + &e. OY elt 
4/ 4// 
at is, Ar >Re i=: + a wu Gi... . ad 


H ' 
condition which will always be fulfilled (Art. 315), when 


u= or <=—, K being the greatest co-efficient of wu. 


A’ 
K+ A! 
Ul. If any number be substituted for x in these expressions, it ts 
‘possible that any two consecutive ones can become zero at the same 
née. 


\ 





\ 
| 


356 ELEMENTS OF ALGEBRA. (CHaP. X. 


For, let Xn—1, Xn» Xntv be any three consecutive expressions. 
Then among equations (3), we shall find 
XE XQ, + Xp. 4; 
from which it appears that, if X,-, and X,, should both become 0 
for a value of x, X,4, would be 0 for the same value; and since 
the equation which follows (4) must be 
$0 7 XntiQn+1 a Aces 
we shall have X,4, = 0 for the same value, and so .on until we 


should find X,= 0, which cannot be; hence, X,-, and X, can- 
not both become 0 for the same value of a. 


III. By an examination of equation (4), we see that if X, be: 
comes 0 for a value of x, X,-, and X, 4, must have contrary 
signs; that is, if any one of the expressions ws reduced to O by 
‘the substitution of a value for x, the preceding and following ones 
will have contrary signs for the same value. 


[V. Let us substitute a+ wu for « in the expressions X and X,, 
and designate by U and U, what they respectively become under 
this supposition. Then (Art. 297), we have 


U2 


U =A + Aw ae Mees 58%, 


es . (5), | 
U,= A, + A4ut+ ACR + &c. 


in which A, A’, A”, dc., are the results obtained by the subsi 
tution of a for «, in X and -its derived polynomials ; and A, A’, &c., 
are similar results derived from X,. If now, a be a root of the 
proposed equation X = 0, then A = 0, and since A’ and A, are 
eac: derived from X,, by the substitution of @ for x, we have 
A’ = A,, and equations (5) become 


pone / 4/ u? 

U =Au+aA mt - (6). 
U,= A’+A’u 4+ &e. 

Now, the arbitrary quantity w may be taken so small that wher 
added to a, it will but insensibly increase it, and when subtracted 
from a, it will but insensibly diminish it; in which cases, the signs 
of the values of U and U, will depend upon the signs of thei 
first terms; that is, they will be alike when w is positive or wher 
a+ u is substituted for v, and unlike when wu is negative or wher 













JHAP. XI.) STURMS’ THEOREM. 357 


1—u is substituted for « Hence, if a number insensibly less than 
me of the real roots of X =0 be substituted for x in X and X,, 
he results will have contrary signs, and if a number insensibly greater 
han this root be substituted, the results will have the same sign. 


a s 


337. Now, let any number as f, algebraically less, that is, nearer 
qual to — om, than any of the real roots of the several equations 


emer AG = Oyun X sh em Os 


ge substituted for x in them, and the signs of the several results 
arranged in order; then, let « be increased by insensible degrees, 
jantil it becomes equal to / the least of all the roots of the equa- 
sions. As there is no root of either of the equations between & and 
h, none of the signs can change while «a is less than h (Art. 311), 
and the number of variations and permanences in the several sets 
of results, will remain the same as in those obtained by the first 
substitution. 


When 2 becomes equal to h, one or more of the expressions X, 
‘X, &c., will reduce to 0. Suppose X,, becomes 0. Then, as by 
the second and third properties above explained, neither X,,_, nor 
AX, can become O at the same time, but must have contrary 
‘signs, it follows that in passing from one to the other (omitting 
XK, = 0), there will be one and only one variation: and since their 
wiges have not changed, one must be the same as, and the other 
contrary to, that of X,, both before and after it becomes 0; hence, 
‘im passing over the three, either just before X,, becomes 0 or just 
after, there is one and only one variation. Therefore, the redue- 
tion of X,, to O neither increases nor diminishes the number of 
variations; and this will evidently be the case, although several 
lof the expressions X,, X,, &c., should become 0 at the same time. 

If « =A should reduce X to 0, then f is the least real root of 
the proposed ‘equation, which root we denote by a; and since by 
the fourth property, just before « becomes equal to a, the signs 
of X and X, are contrary, giving a variation, and just after pass- 
ing it (before « becomes equal to a root of X,= 0), the signs are 
the same, giving a permanence instead, it follows that in passing 
(this root a variation is lost. In the same way, increasing wv by 
‘insensible degrees from x= a-+ wu until we reach the root of 
‘X = 0 next in order, it is plain that no variation will be lost or 
gained in passing any of the roots of the other equations, but that 


358 ELEMENTS OF ALGEBRA. [CHAP. XT 


in passing this root, for the same reason as before, another varia- 
tion will be lost, and so on for each real root between 4 and the 
number last substituted, as g, a variation will be lost until x has 
been increased beyond the greatest real root, when no more can be 
lost or gained. Hence, the eacess of the number of variations ob- 
tained by the substitution of k over those obtained by the substi- 
tution of g, will be equal to the number of real roots comprised 

It is evident that the same course of reasoning will apply when 
we commence with any number p, whether less than all the roots 
or not, and gradually increase x until it equals any other number g. 
The fact enunciated in Art. 335 is therefore established. 


between & and g. 


338. In seeking the number of roots comprised between p and g, 
should either p or g reduce any of the expressions X,, X,, We. 
to 0, the result will not be affected by their omission, since the 
number of variations will be the same. 

Should p reduce X to 0, then p is a root, but not one of those 
sought; and as the substitution of p+ u will give X and X, the 
same sign, the number of variations to be counted will not be 
affected by the omission of X = 0. 

Should g reduce X to 0, then qg is also a root; and as the 
substitution of g —u will give X and X, contrary signs, one varia- 
tion must be counted in passing from X to X,. 


339. If in the application of the preceding principles, we ob 
serve that any one of the expressions X,, X,... &c., X,, for in 
stance, will preserve the same sign for all values of « in passin 
from p to qg, inclusive, it will be unnecessary to use the succeed 
ing expressions, or even to deduce them. For, as X, preserves 
the same sign during the successive substitutions, it is plain the 
the same number of variations will be lost among the expression: 
X, X,, &c.... ending with X,, as among all including X,. When 
ever then, in the course of the division, it is found that by placing 
any of the remainders equal to 0, an equation is obtained witl 
imaginary roots only (Art. 325), it will be useless to obtain any 
of the succeeding remainders. This principle will be found very 
useful in the solution of numerical examples. 


340. As all the real roots of the proposed equation are neces: 
sarily included between — m and + @, we may, by ascertaining 
| : 








- CHAP. XI.) STURMS’ THEOREM. 359 


_the number of variations lost by the substitution of these, in suc- 
cession, in the expressions X, X,...X,,.. &c., readily determine 
the total number of such roots. It should be observed, that it will 
| be only necessary to make these substitutions in the first terms 
, of each of the expressions, as in this case the sign of the term 
_ will determine that of the entire expression (Art. 315). 


341. Having thus obtained the total number of real roots, we 
“may ascertain their places by substituting for 2, in succession, 
the values 0, 1, 2, 3, &c., until we find an entire number which 
gives the same number of variations as + o. This will be the 
} smallest superior limit of the positive roots in entire numbers. 

_ Then substitute 0, —1, —2, &c:, until a negative number is 
_ obtained which gives the same number of variations as —@ ‘This 
will be, numerically, the smallest superior limit of the negative 
roots in entire numbers. Now, by commencing with this limit and 
observing the number of variations lost in passing from each num- 
. ber to the next in order, we shall discover how many roots are 
included between each two of the consecutive numbers used, and 
thus, of course, know the entire part of each root. The decimal 
part may then be sought by some of the known methods of ap- 
| proximation. 


EXAMPLES. 

1. Let 823 — 62 —-1=0= X. 

The first-derived polynomial (Art. 297), is 
24x? — 6, 


and since we may omit the positive factor 6, without affecting the 
| sign, we may write 
I 4a7—1= hon 

Dividing X by X,, we obtain for the first remainder, — 4x — 1. 
Changing its sign, we have 


424+1= X,. 
Multiplying X, by the positive number 4, and then dividing by 
X,, we obtain the second remainder — 3; and by changing its 
k sign 
+3= X;. 


The expressions to be used are then 
X = 8x3 —Ge—1, X,= 40? —1, X,=4¢+ 1, X,= +3. 


360 ELEMENTS OF ALGEBRA. {CHAP. XI. 


Substituting — o and then + @, we obtain the two ae 
arrangements of signs: 


—+—+...... 3 variations, 


a-ha taal . 0 54 
There are then three real roots. 
If now, in the same expressions we substituile 0 and + 1, 
and then 0 and —1, for 2, we shall obtain the three following © 
arrangements : 


For e—=+l1 +++ 4 0 variations, 

se Ca 0 oo ae + ae 1 

« | ne ehdaveetlt ae 3 nuit 

As «= +1 gives the same number of variations as + o,- and 
x= —1 gives the same as —o, +1 and — 1 are the smallest 
limits in entire numbers. In passing from — 1 to 0, two variations 
are lost, and in passing from 0 to + 1, one variation is lost; hence, 


there are two negative roots between — 1 and 0, and one positive 
root between 0 and +1. 


a Let 224 — 1322 + 107 — 19 — O. 
If we deduce X, X,, and X,, we have the three expressions 
X = Q2* — 1322. 10 
X= 403 — 1324555 
X, — 13220— 15apee 38. 
If we place X,= 0, we shall find that both of the roots of 
the resulting equation are imaginary; hence, X, will be positive 


for all values of x (Art. 325). It is then useless to seek for “4 
and X;,. 

By the substitution of — ao and +o in X, X,, and X,, we 
obtain for the first, ¢wo variations, and for the second none; hence, 
there are two real and two imaginary roots in the proposed equation 


3. Let x3 — 5x? + 8 —1 = 0. 
4. e*— 2 — 322? + a? -—e —3 = 0. 
2. x — 23 4.1 =0. 


Discuss each of the above equations. 


; CHAP. XI.] RESOLUTION OF CUBIC EQUATIONS. 361 


: 


Young’s Method of resolving Cubic Equations. 
1B 34a. Every numerical equation of the third degree may be re- 


duced to the form 
x3 + be? + cr = N Cy; 


in which 8, c, and N, are known numbers. 

_ Since this equation will have at least one real root (Art. 277), 
_let us find, either by Sturms’ theorem or by trial, two consecutive 
‘ numbers, either integral or decimal, which being substituted for a, 
will give results with different signs. We then know that one of: 
' the values of « will lie between them (Art. 311), and consequently, 
that the smallest number will be the first figure of the required 
| root. 

_ Let us designate this figure by r. Now, if we neglect the re- 
maining figures of the root, and regard r as the approximate value 
of x, we shall have 








N 
r+ br te 


Having found r, denote the remaining part of the root by 7; then 


rs + br? + er = N; whence, r = 


tore 7 yy, 
Substituting this value of x in the given equation, we have 
Paar ary + ary? + 7 
bu? = br? + 2bry + by? i 
Gx Seen +-/cy 
and by adding and arranging with reference to y, 
y> + (37 + 5) y? + (372+ 2r4+ c)y+ (2+ br? + cr) = N. 
But we may simplify the form of this equation, by making 
Y=3r4+b ¢& =3r42r+c, Wa N— (P+ br? + cr); 
which will give y+ Wy?+ “y= N (2). 
The form of this equation is entirely similar to that of the given 


equation; and if we denote by s the first figure of y and make 
the same supposition as before, we shall have 
8+ 0s? +¢s=N’; 

IV’ 


7 whence, a, 2LWste 


~~ * 
eo 


362 ELEMENTS OF ALGEBRA. [CHAP. X1, 


Supposing the value of s found; denote by z the remaining 
figures of the root: then 


y =5s4+2, 
aud Pos + 3s2z + 382? + 23, 
b’y? = b/s? + 2h’sz + b’2?, 
CY mri C Sie Cee 


By adding and arranging with reference to z, 
23 + (3s + b’) 2? + (3s? + 2h’s + ec’) z + (53 + Ws? + e's) = N’. 
But, by making b” = 3s + U, 
ec’ = 3s? + 20s +c, 
N”’ = N — (s3 + 0s? + e’s), 
the equation becomes 
23+ 6/22 + ¢’z = N” (3). . 
If we denote by ¢ the first figure of z, and make the same 
supposition as before, we shall have 
RETO ee ees 
NY” 
4 Ott 
If we designate by wu the first figure of 4, we should find, by a 
process similar to the above, 


whence, t= 


N’” 
. Use a Bee om? 
and in a similar manner we may find the algebraic expression for 
any succeeding figure of the root. 
343. It is now required to put these algebraic expressions under 
such a form as will indicate the best practical rule for performing 
the arithmetical operations. For this purpose, let us bring the for- 





mulas together. We have 


N ee ne 

Neer br 4b ~ r(r +6) +e’ 

Th aah Wes: cite ~ s(s + 3r-+ 6) + 37? + br $c’ 

pee ates! why plant Oi 

Pb tte tlt +3 (r+ 5s) +d] + 3s? 4+ 20s 4 c”’ 
NY” NY” 





PLY ut ee” ulut3(rt std +b) 430+ 04+ 
&c., &c., &e 














CHAP. XI.] ° RESOLUTION OF CUBIC EQUATIONS. 363 

The value of 7+ being found, and c a known number, the de- 
nominator r(r + b)+c will be known. This forms the first di- 
visor, and dividing N by it, the first figure of the quotient will be 
7, as before found. Multiplying the divisor by r and subtracting 
the product from N, we obtain N’, the second dividend. 

It will be seen that the three right-hand terms in each denomin- 
ator, are formed from the preceding figure of the root. ‘These 
make trial divisors for each figure of the root after the first. 

Having found N’, we form its trial divisor and then see how 
often it is contained in N’, which gives s. We then form the 
complete divisor which we multiply by s, and subtract the result 
from N’, which gives VW”. We then form its trial divisor, find the 
figure t, after which we find the complete divisor for ¢, and then 
multiply it by the quotient figure ¢ and subtract the result from NV’, 
giving N’”; and similarly for all the following figures of the root. 


344, By examining the table of Example I, on the next page, 
and comparing it with the formulas, we see, that if under any divi- 
sor we write the square of the figure of the root which the divisor 
determines, and then add it to the two numbers directly above, their 
sum will be the érial divisor for the next figure of the root. Hence, 
we have the following 

RULE. 

I. Write down c, the co-efficient of x, and on the same ne, to the 
right, place the known number N, and set in the quotient il gure 
of the root found by trial. ‘ 

Il. To this figure of the root add b, wo co-efficient of x, and 
then multiply the sum by: the figure of the root, and add the product 
fo c, and the sum will be the first divisor, which is then to be 

multiplied by the quotient figure and the product subtracted from N. 

Ill. Under the first divisor write the square of the first figure of 
the root, and then add it to the last two sums, and the result will 
be the trial divisor for the next figure of the root. 

IV. Having found the neat figure of the root, add to it three 
times the figures of the root already found, and also the co-efficient 
b: then multiply the sum by the figure of the root and add the prod- 
uct to the trial divisor, and the sum will be the entire divisor, which 
must then be multiplied by the figure of the root, and the product 
subtracted from the last dividend. The process for determining other 
figures of the root is entirely similar. 


(CHAR, Xe, 


7 


ALGEBRA, 


ELEMENTS OF 


364 


= ° ~ 
I | l 
6SSL = 29 4ec*8 $F ct 
uonyenbs oy} UI £ Jo on[eA oUO pUly *[ 


“ATH NVXG 


‘9p 


‘pusplaip YG =,,,,N = CLEP8PPEO’ O 
“pusyerqns Yip =2,,9+22,,9+—eN= CZOLZ8IIE’ O 





Poe PE WP = 4. GIE9GE * 0 

‘pueyenqns pe =7,9+27,9+9= gg96ec'T 

“Puoplsp Pt.=— NS 966° T 
‘pucyeqns py =s-+,8,9+.s= POL * Zz 
Usp ae Ne 6° &% 
‘puoyenqns is— =w9+,09+,u= 2G 

‘S]001 oY} JO 900 Jo ‘x= +ec4cl6 ‘CL=N 
00 a 


“GGSC9E * ZO 
| GcE9L0° 0 
G68G ° G9 
vO00 ° 0 
vr86 “19 


VHOE © 





aoe ee 


91” 
94° GG 
9L°S 


st uonenba 


| 


I 


| 


I 


ROSY Ee 

USAIS OY} JO SJOOI OY} JO auO ‘aouezT 
Sn 

wt ++s+4)e+n]n= rostarp yp 
= fa+@+s+s3)e + a}n 
WP = ATLAS E gE = IOSIaIp [vl pE 
= 2? 
P+lat (¢ +4) ¢ +4] 2 = sosttp pe 
[9 +(s+ 4) e+Fa]2 
P= P+ SAB + SE = IOSIAIp [eI pZ 


= 2s 
- P+(q+ 4¢ + s)s= Iostaip pz 
- = (Q+4g+5)s 

- P~=9+ 19% + 2G = JIOSIAIP [PIN IST 
= = et 


-9+ (9 + 4) 4 = sostarp 48] 
(g@t2e=(¢+-)4 








- CHAP. XI.] =| RESOLUTION OF CURIC EQUATIONS. 365 


Remarx.—The operations in the example of the table, are all 
performed according to the directions of the rule; but more deci- 
mal places have been used in the dividends and divisors, in the 
latter part of the work, than were necessary. Had we admitted 
but three places of decimals in the dividends, and rejected all other 
places to the right, as fast as they occurred, we should still have 
had the root equally true to at least four places of decimals. But 
' since the figures of the root are decimals, it follows that if the num- 
_ ber of decimal places in the dividend does not exceed three, the 
_ decimal! places in the corresponding divisor should not exceed two ; 
_ and for every succeeding figure found in the root, one place may 
be struck off from the right of the divisor. 

' After finding a certain number of figures of the root, it will 
occur that the numbers to be added to the divisors will fall among 
the rejected figures, after which the remaining figures of the root 
will be found by simple division. It should be observed, however, 
. that when places are rejected from the divisor, that whatever would 
have been carried had the complete multiplication been performed, 
is still to be carried to increase the last figure retained; and when- 
ever the left-hand figure of those rejected, either in the dividend or 
divisor, exceeds 4, the last figure retained is to be increased by 1. 

The following is the last example, wrought on the principle of 
admitting but three places of decimals into the dividends. The 
rejected figures, both in the dividends and divisors, are placed a 
little to the right. 





6 75 09 |2.4257 4 
20 52 
26 23.9 
4 | 22 . 304 | 
50 1.596 
5.76 1.240 | 688 
55.76 B00" (, 312 
. 16 812 | 827625 
61.68 .044 | 484375 
.30| 44 043 | 
62.0 | 844 001. 
0 | 004 , 
62.3 | 892 
1 | 76325 





366 ELEMENTS OF ALGEBRA, [CHAP. XI, 


2. Find one root of the equation a? + «2 = 500. 

This equation is the same as a? +- 2?+ 0x2 = 500; hence 6 =1 
¢= 0, and. Ni = 500. 

The first figure of the root found by trial, is 7. 








0 500 |7 . 61727975, &c., = x. 
56 392 
56 108 Bees 
49 104 . 736 ag eB 
161 3.264 ee 
Ee Maes te 
1Bie56 1. 887181 ase 
174 . 56 1 . 376819 1 erie 9 Bp 
36 1 . 323862 <5 ie : 
188.48 . 052957 Ges oa 
4 i 
2381 037859 = ST 
188.7181 . 015098 aS HS ay 
__-0001 013251, Fig on 8 
188 . 9563 "001847 iPass ia 
. 1669 . 001704 Ml Re ei i 
SS iter Gouin) pa eelbainet Ss Tos ee wm. FS 
189 . 123)2 - 000143 <'3 0 2 3 * 
189 . 290 WOCHeS Bre 8 we 
. 005 . 000010 pig eh, 
18/9 . {2]9]5 - 900009 





3. Find one root of the equation «#3 — 17x? + 54” — 350. 
Ans. x = 14.954, &c. 


4. Fine ~e root of the equation x? + 2x? + 3x = 13089030. 
Ans, 2 = 235: 

5. Find one root o: ‘e equation #3 + 222 — 23% = 70. 
Ans. x = 5.1345, &c. 


Remarx.—In the preceding soluu.:°s only one root has been 
obtained, yet the others may be found witn o-ual facility, by find- 
ing by trial the first figure in each and then preceeding by the 
tule already given. There is, however, a shorter metis! for de- 
termining the remaining roots. 

“Subtract the root found, taken with a contrary sign, from the co- 
efficient of the second term of the given equation, and denote the 
remainder by a. Divide the absolute term by the root found, and j 
denote the quotient by 6; then will the roots of the equation ory 

“vtar+hb=0 ae 


be the two remaining roots of the given equation. 








‘CHAP. XI.] RESOLUTION OF CUBIC EQUATIONS. 367 


Method of resoluing Higher Equations. 


| 345. The general method of resolving cubic equations, has been 
| explained in Art. 342. We shall now add from Young’s Algebra, 
‘the method of resolving equations of a higher degree. It has 
not been thought best to give the general investigation, but merely 
‘to add, for the solution. of an equation of any degree, the follow- 
‘ing general 


RULE. “f 


| I. Transpose the absolute term to the second member of the equation. 
| Then, beginning with the co-efficient of the first term, arrange the co- 
efficients of the first member in a row, placing the absolute term to the 
right. 


If. Having found the first figure of the root, multiply it by the first 
|| co-efficient and add the product to the second co-efficient; then multiply 
|| the sum by the same figure of the root and add the product to the third 
co-efficient; and soon to the last co-efficient: the last sum will be the first 
divisor, which multiply by the figure of the root and subtract the product 
from the absolute term: the result will be the second dividend. 


III. Perform the same operations on the first co-efficient and the set of 
sums found, as was performed with the co-efficients, and the last sum will 
be the TRIAL Divisor for the second figure of the root. Then perform 
the same operations on the first co-efficient and the second set of sums, only 
stop in the column of the last co-efficient but one. Repeat the same op- 
eration on the first co-efficient and the last set of sums, but stop in the 
next left-hand column, and so on until you stop in the column of the sec- 
| ond co-efficient. | 


IV. Then find from the trial divisor the second figure of the root, 
taking care that it be not too large. Take this second figure, and per- 
form with it on the first co-efficient and the last set of sums the same 
| operations as were performed on the co-efficients with the first figure of 

the root and the sum; in the last column will be the second divisor, which 
“multiply by the second figure of the root and subtract the product from 
the second dividend. 





V. The next trial divisor, the next figure of the root, and the true 
divisor, are found by the principles already expluined, and the places 
| of figures in the root may be carried as far as necessary. 


348 ELEMENTS OF ALGEBRA. [CHAP. XI. 


EXAMPLES. 


1. Find the root of the equation a — 3x? + 75x” = 10000. 








Operation. 
Gan. 238 75 10000 |9. 8860027, &c., =z 
9 81 702 6993 
Oy 78 777 3007 
9 162 2160 2677 . 5616 
18 240 2937 329 . 4384 
9 243 409 . 952 306 . 1662 
27 483 3346 . 952 23 . 2722 
9 29.44 434 . 016 23 .2616 
36.8 512.44 -.3780.968 106 
8. 30.08 46.110 78 
875 Gr) 54055 m. 3627 . 07/8 28 
8 30.72 46 . 36 Q7 
S804) BY) Bat. 88739. 44 4 
8 214 3.50 
3|9.|2 576.3|8 3876. 9|4 
epee 5 
579 . |5 3880 . 4 
3 
a 
5/8|3 


Remarkx.—The work, in the example, ha been contracted by 
omitting or cutting off decimal places, as in the operations for the — 
cube root, and in equations higher than the third degree, the con 


_ tractions may be begun after the first decimal place of the root is 


gound, 


_ By using one period of four decimal places, the root has been 
Hfonnd to eight places of figures. Another period of four places, 
that is, by beginning the contractions later, we should have found 
four additional places, or a = 9 . 88600270094. y 


THE END. : 

2 . 
f. rd <* of A Pa -_ 

de Lt A) ah &7 


: 
: 





7 = 
ee 





ints Uh A ime 
' Ae Oe bh 
fa iy, wei) rs : 
4 ay Me rman; @ Se b. AS 
ies va y i 
va 
ih YI 


OP sb j , 
iy hi ae a ‘ 
a s 
5 ae ay, 
(Dy m 
7, etal j t ‘ 
a ae ee 
ST hd ae ee 
Var Tee i) Gan) yi 
ain ts f Todt 
4 - ree ‘4 5) ans 
4 é id 4 
Un, ? . 
1 
LX / 
“ re 
, J 
by! 
44 * 
) 
? 
iy 
j 
¥ ‘s 
} 
1 j f 
‘i 
to: f 
mi 
Lf . 
> / in { B, vit 
i Vd): 
if ¥. | 
Vas : 
oi i F » 2 
Soi, Ak) ee 
i f vi 
iy ‘ a 
5 a: 
a 
j 
| f J Py j 
' 
i] 
ms we 
. - 
; , 4 4 j M 
¢ } 1 
bie. ‘ ; 7h 
a Von lek | 
ny 7 Ua p) 
aL \ 4 \ 
i Oy Se Ce Bee 
ee Orn 
Po Co et eo 
‘pi ae Cae ee ae 
wi => ¢ 5 sar F 
ee <1 \!Gmoe BR raat: 
bret +4 yn 
ie vies: 
a4 MA yt ass 
21 San ATA 





Paes ] 
TRA. 
hag ei 


i! 


eaied 





1 
jee; 
ie 





cnet 


= 





OCT 2 1972 








